How much change in Earth's orbital distance to change average temperature by 1°C? How much closer/farther would the Earth need to be to/from the Sun to effect a $1 \sideset{^{\circ}}{}{\mathrm{C}}$ increase/decrease in average temperature?
 A: If you want to do this properly you have to take into account the Earth system, which is, to put it mildly, extremely complicated, and probably off-topic for Physics SE.
But being physicists we like oversimple approximations, especially when they give answers which are approximately correct.  One such is to treat a planet as a perfect black body, and use this to calculate how hot it should be at a given distance from the Sun.
First of all, given the known top-of-atmosphere flux of power from the Sun, $S$, we can, by integrating over a sphere whose radius, $R$, is the Earth's orbital radius (semimajor axis), compute the total power output of the Sun:
$$P_S = 4\pi R^2 S$$
We know $S \approx 1360\,\mathrm{Wm^{-2}}$ and $R \approx 1.50\times 10^{11}\,\mathrm{m}$ and this gives $P_S \approx 3.85\times 10^{26}\,\mathrm{W}$.
Now we can use the black-body formula and the above expression for the flux in terms of $P_S$ and remember that only half of the planet is illuminated to compute the predicted temperature of a perfect black-body planet:
$$T = \left(\frac{P_S}{16\pi\sigma R^2}\right)^{1/4}$$
And we can check this for Earth, and we get $T \approx 278\,\mathrm{K}$: this is too cold but is clearly in the ballpark (it is entertaining to do this for Venus, where the answer is not in the ballpark).  So the approximation is not hopeless.
We can invert the above expression to get $R$ as a function of $T$:
$$R =\frac{1}{T^2}\sqrt{\frac{P_S}{16\pi\sigma}}$$
And then we can just plug in numbers to this: for an increase of $\Delta T = 1\,\mathrm{K}$ the $\Delta R \approx -1\times 10^9\,\mathrm{m}$, and for a similar decrease, $\Delta R \approx 1\times 10^9\,\mathrm{m}$: this is about a million km, or about $1/138\,\mathrm{au}$: it's rather under one percent of $R$ in other words.
I will emphasize once again: this is insanely oversimplified: you can't treat planets with atmospheres, life, oceans &c this way and expect to get reasonable answers: look, again, at Venus (which has only some of these things) for an example.
(And, just in case, no, climate change is not being caused by the Earth moving closer to the Sun!)
A: We can make a very crude order of magnitude of the required extra energy $δE$ as such :
the mass of atmosphere,  oceans, and polar ice, are respectively about 50, 1500 and 30 Ekg ( 1 Ekg = one exakilogram = $10^{18}$ kg). 
Assuming specific heat capacities or 1, 4, and 2 kJ/K/kg , neglecting the temperature change in the earth itself, an increase of temperature by 1K will correspond to  an energy of $δE\sim (50×1 +  1500× 4 +30× 2) 10^{21}  \sim 6 ×10^{24}$ J. To the which should be added the enthalpy of fusion of one part of polar ices. Assuming 1/3 of ice melt, ant with the latent heat of ~300 kJ/kg, this results in $\sim 30× 300/3 10^{21} = 3 10^{24}$ J, for a total of about $δE~10^{25}$ J. Notice that this estimate neglects the heating of the rocks, and the melting and heating of the permafrost, which, among others, could significantly increase the total required energy.
On the other side, the average power flow received from the Sun is presently of about $1 \mathrm{kW/m^2}$, and multiplying by the cross section $π R_t^2$, one get a total power of about $170$ PW ( 1 PW = one petawatt = $10^{15}$ W). The other sources of heating can be neglected in this simple calculation
Now, can we convert the above calculated energy into a power ?
 This would require a sophisticated model,  which will be very sensitive to climate change, and that can't be undertaken here.  But let's us make a few comment about it. One key parameter is the albedo coefficient which  measures  the fraction of power which is immediately reflected toward space without heating the earth. It is presently of about 35%, but will decrease significantly if the surface covered by clouds or by ice is reduced. One part of the absorbed energy is re-emitted, and the relationship between radiated power and an temperature can be modeled.
But the problem is in fact much more complicated, as a significant part of the incoming radiation is absorbed by atmosphere (see Wikipedia). Similarly a part of the re-emitted radiation is absorbed or reflected back y the atmosphere and clouds : this part corresponds to the so called greenhouse effect, evaluated presently to about 5% of the incoming power. It  was originally a benefic natural effect, but has dramatically increased by human activities and will increase again a lot, because heating the earth, atmosphere and ocean, displaces many subtle equilibrium. Among others, we can neither know with sufficient precision how much the excess CO2 will be captured by ocean, nor how much methane will be released in atmosphere by the melting of permafrost.
Any way, the short answer is that we can not convert this energy into power. 
But this is not required. If climate change is ignored, and the present parameters of earth are used, and one assume that the present energy's balance is at equilibrium. In this restrictive context, we can use the relationship between radiated power and temperature stated for a black body by the Stefan law ($P \propto T^4$). Earth is clearly not a black body, and the "radiated power" is not the real power sent back to space, but if "all else being equal" (especially same albedo, same greenhouse effect, etc. which is completely unrealistic) the unknown coefficient must be the same in the two considered situations.
Hence the calculation is straightforward: the relative increase of power is about $δP/P = 4 δT / T$, and as the power flow is proportional to $1/D^2$, $δP =-2 δD/D$.
As the black body model of earth (once included all side effect) corresponds to an effective temperature close to 300K, we get : 
$$ \frac{δD}{D} \approx -2 \frac{δT}{T} \approx  -1/150 $$
As $D=1 \mathrm{UA}=150\times 10^6$ km, we get $δD \approx -10^6$ km.
This value could seem very large, but considering the eccentricity of the Earth's orbit $e \approx 0.016$, we observe that it is only $50$ times larger than the difference of the major axis and minor axis of the orbit. 
Finally the energy can be used to evaluate the the characteristic time for reaching the new equilibrium :  $τ_c= δE/δP = 4 δE/P x T/δE \sim 10$ days, which is again unrealistic.
