How much solar energy do you entrain for every unit of energy you use, when that
  energy is derived from fossil fuels?

When we use fuels to liberate heat, we liberate CO₂, and this traps more heat in the
earth's atmosphere.  How much more heat?  Let's estimate it:

Suppose you use a small space heater (1000 Watts, W) for 1 hour.  A Watt is 1 Joule (J)
per second, so over the hour you used 3.6 million J (3.6x10⁶ J, in scientific notation, which
is a lot easier to use for big numbers; read it as "3 point 6 times 10 to the 6th", with the
last phase meaning 10 raised to the 6th power, or multiplied by itself 6 times, to give a
million). For now, we'll assume that's all the fuel used- this was a heater directly using fossil
fuel, say, kerosene.

How much CO₂ did that put into the air, and how much solar energy did that help
retain at the earth's surface while that CO₂ stayed in the atmosphere?

The amount of CO₂ can be estimated from the energy content of a typical fossil fuel-
about 40,000 J per gram (4x10⁴ J / g) - and the CO₂ produced per g of fuel.
Taking that first number, we have that
   Mass of fuel used = (energy used)
                                 ---------------
                                (energy delivered per mass of fuel used)

                              = (3.6x10⁶ J) / (4x10⁴ J / g)
                              = 90 g   (OK, about 3.2 ounces, for the metrically challenged)

How much CO₂ is liberated to the air in burning 90 g of fuel?  The average chemical
composition of liquid fuels is close to CH₂ (say, C₁₀H₂₀, for the full molecule).
Chemists work in units of moles,; 1 mole is a fixed number of molecules (Avogadro's
number, 6.02x10²³, a huge number), and it equals the mass in grams of all the
chemical elements, multiplied by their atomic masses.  Hydrogen (H) has a mass of 1
(OK, 1.008, really, with some normal isotopes present), and carbon (C) has a mass of 12.
So, CH₂ has a molecular weight of 14; every 14 g of the fuel contains 1 mole of C.
So, our sample of 90 g of fuel contains
      n = number of moles of C = (mass of fuel used) / (mass of fuel per mole)
         = 90 / 14
         = 6.4, with appropriate rounding

Oops - only about 40% of the CO₂ we inject directly into the air by burning fuels
stays in the air.  Right now, the rest is absorbed by the ocean or by plants on the
land.  So, only about 0.4 x 6.4 = 2.56 moles of CO₂ ended up in the air in the short term
when we used that heater.

How much solar energy does that help trap at the earth's surface?  Every mole or every
molecule of CO2 acts the same, so let's compute the total amount of CO₂ added by
human activities since the Industrial Revolution, and relate it to the change in the earth's
temperature.  We can then convert the change in temperature to the amount of heat
that has to be radiated back to the earth from our layer of greenhouse gases.

How much has the CO₂ content of the earth's atmosphere increased?  As a concentration,
it has gone from 280 parts per million (ppm; our best estimate, from ice cores in glaciers
that have preserved samples of the ancient air) to about 370 ppm now (measured by
a worldwide network of samplers, with the one on Mauna Loa, Hawaii, keeping the
longest record, since about 1957).  So, we've added about 90 ppm.  In the same time,
the average increase in temperature worldwide is about 0.5 degrees Celsius (0.5oC),
or about 0.9 degrees Fahrenheit).  This is the best estimate of the part of climate change
due to our adding CO₂ to the air.

An increase of 90 ppm must be mulitplied by the total number of moles of air to see
how many moles of CO₂ we have added.  Each square meter of the earth's surface has
about 10,000 kg of air above it (you can work this out from the air pressure and the
gravitational acceleration constant).  Now, the earth's surface is 4 pi times the square of
the earth's radius, which is about 6400 km (about 4000 miles).  To cut to the chase, the
area is about 53 billion hectares or 5.3x10¹⁴ m2 (square meters).  So, the earth's
atmosphere contains about (10,000 kg / m2 ) x (5.3 x 10¹⁴ m2) = 5.3x10¹⁸ kg of air.
The average molecular weight of air is between that of nitrogen gas (N₂; about 28 g per
mole) and that of oxygen gas (O₂; about 32 g per mole); given that there's more N₂ than
O₂, the average molecular weight is close to that of N2, or about 29 g per mole.  Let's
put this in kg, as 0.029 kg per mole.  Then, the number of moles of air is about
   n_air = (mass of air) / (molecular mass per mole of air)
          = (5.3x10¹⁸ kg) / (0.029 kg /mol)
          = 183x10¹⁸ mol
          = 1.83x10²⁰ mol  (the abbreviation of "mole" is "mol")

How much of this is our added CO₂?  We have to multiply this by 90 ppm, or
90x10^-6 (0.000090).  We get the figure
    n_addedCO2 = 1.83x1020 mol x 9x10-5
                        = 1.65x1016 mol as CO₂

How much solar energy did each mole of CO₂ trap?  We have to know how the energy
that's radiated is related to temperature.  All bodies above absolute zero in temperature
radiate energy as thermal infrared.  There's a lot of information about this, but for us, the
important point is that the amount of radiation increases as the fourth power of the
body's temperature.  So, if the absolute temperature doubles, the amount of radiation
increases by a factor of 24 = 16.  Absolute temperature is that temperature measured
relative to absolute zero, which is -273oC or -459oF.  So, an object at 20oC is at an
absolute temperature of 293, in units we call Kelvin to distinguish them from Celsius on
a different scale (from Fahrenheit, the change in name is to "Rankine").  There is a
fixed constant of proportionality, the Stefan-Boltzman constant, symbolized by the
character sigma (which I can't readily get in this composer package).  Then, the
rate (I) at which thermal infrared energy is radiated from a given area of an object at
absolute temperature Tabs is 
    I = (sigma) T_abs⁴
We can add a small factor, the emissivity, to account for the fact that real objects don't
act as perfectly black bodies ("black"has almost nothing to do with color in visible light
note).  For the earth's surface, the factor is close to 1 (about 0.96), so we'll ignore
it for now.  To a first approximation, then, the earth's surface has an average temperature
of 15oC or 288 K; it then radiates thermal infrared at a rate
   I = 5.67x10^-8 (W/m2)/(K⁴) x  (288 K)⁴
     = 390 W / m²
An increase of 0.5oC changes this to the same expression, but with 288 replaced by 288.5.
The increase in I is computed as 2.7 W / m². (Note: this calculation is not the same
as the calculation of radiative forcing, and it uses our transient increase in global temperature,
not our final or equilibrium increase, which is likely to be twice as much after accounting
for the greater absorption of sunlight when glaciers and sea ice melt, etc.)

Over the whole surface of the earth, our gazillions of moles of CO₂ (OK, 1.6x1016 moles)
caused an additional heat retention that resulted in the radiation of
    E = 2.7 W / m² x (area of earth's surface)
       = 2.7 W / m² x 5.3 x 1014 m²
       = 14.3x1014 W - a lot, of course
This was contributed by 1.6x10¹⁶ moles of CO₂, so each mole of CO₂ helped trap
an amount of radiation
    E' = (total radiation increment) / (number of moles of CO₂)
        = (14.3x10¹⁴W) / (1.6x10¹⁶ mol)
        =  0.09 W / mol

Our piddling 2.56 mol added 2.56 x 0.09 = 0.23 W to the earth's surface radiation....
but, as the difference in units tells us (Joules, vs. Watts = Joules per second), we
are comparing apples and oranges.  How long did the 0.23 W addition persist?
About as long as the CO₂ stays in the air!  This is about 50 years for the average
CO₂ molecule.  Each year is (with rounding) 365 days x 24 hours per day x 3600
seconds per hour = 31 million seconds (3.1x10⁷ s).  Then, the total heat that was
trapped by the amount of CO₂ we added is about
    H = (0.23 Joules / s) x (3.1x10⁷ s/yr) x 50 yr
        = 3.6 x 10⁸ J  or 360 million Joules
This is 100 times larger that the original heat liberated in burning the fuel!

The story doesn't quite end there.  If the burner were electric, the fuel was burned
at the power plant, with only about 35% having been converted into electric power.
Only about 90% of that got to you, after losses in the power lines.  So, the net
efficiency is about 31.5%.  The amount of fuel actually burned, and thus, the
amount of CO₂ staying in the air, is larger yet by a factor of 1/(0.315), which is
about 3.2.  The fuel did not even get to the power plant itself without fuel being
burned to mine and transport the fuel, and to make the equipment an the
buildings.  These efforts inflate the amount of fuel used by another 20% or so,
so we're up to 3.2 x 1.2 = 3.84 times as much CO₂ as before.  The amount of
heat trapped by all the fuels we used is almost 400 times as large as the heat we
got from the heater!  We can't use fuels without looking at the long-term
consequences.

Some other interesting aspects:

The amount of extra thermal radiation emitted - at the top of the atmosphere - from fuel use
and CO₂ retention in the air is called "radiative forcing" (see above) by climate scientists
(climatologists and climate modellers). Radiative forcing does not come only from adding CO₂;
we also change cloudiness, as well as the amount of aerosols in the air.  Aerosols, such as smog
as well as ordinary soot, can add to or subtract from the final radiative forcing.
The best estimates of total radiative forcing from human-induced change (fuel
burning, forest clearing, etc.) is about 1.2 W / m2.  This is somewhat less than half
of what we computed above.