Do you pay more for gas when the day is warmer? Found this at the gas station yesteday - got me thinking...

 A: My belief is that this refers to the density effect alone.  Obviously you will get less gas on a hot day since it will be lower density at that temperature.
This is separate from thermodynamic factors.  I would venture an educated guess that ICEs will be less efficient on hot days, and I would suspect that this factor would be greater than the density difference.  Nonetheless, you could still fill up on a hot day and later drive on a cold day (or vica versa, as you might do after reading this answer).  As the temperature changes, the level in your tank will change with it.
As you drive your car, you adjust the throttle to get the needed torque regardless of the density of the gas, so the total mass of fuel is a metric that is much more directly relevant to satisfying your transportation needs.
I can find one reference that gives different densities of crude oils at different temperatures.  Here, I find the points $(77^{\circ} F, 1.014 g/cm^3)$ and $(113^{\circ}, 1.002 g/cm^3)$.  Crude oil is different from gasoline, but mostly because it contains a wider variety of Carbon chain lengths, and I think the density sensitivity to temperature should be on the same magnitude.  I will report a finding that a percentage change in density for a degree Fahrenheit is about:
$$-0.03 \frac{\text{%}}{^{\circ} F}$$
So if the temperature difference between seasons is a fairly typical $30^{\circ} F$ difference, then we're looking at a difference of about $1 \text{%}$ in price.  Obviously that comes out to multiple cents per gallon.  Anecdotally, I know quite a few Americans who would switch gas stations over smaller amounts.  It would also matter a great deal to the gas station's margin.

Again, Wolfram alpha has proven to be useful beyond my expectations.  Here is the exact formula to find the density percent difference for octane.  Octane is fairly representative (but not exactly) of a fuel with a 100 octane rating.  This gives $-0.06395 \text{%}/^{\circ}F$, which is twice what I had for crude oil.  So we are looking at closer to a $2 \text{%}$ difference in price for a $30^{\circ}F$ difference in temperature.

Another edit: Basis in equations to show that what I did is legit.  I'll use price, some constant, and volume respectively in the first equation.
$$P=\alpha V$$
Next equation has mass, density, and volume respectively.
$$M = \rho V$$
I'll combine these to get the final equation that makes the point.
$$P/M = \frac{\alpha }{\rho} \approx \frac{\alpha}{\rho_0} - \rho \frac{\alpha}{\rho_0^2}$$
$$\Delta (P/M) = - \frac{\Delta \rho \frac{\alpha}{\rho_0^2} }{\frac{\alpha}{\rho_0}} = -\frac{\Delta \rho}{\rho_0} = - \frac{d\rho}{dT} \Delta T$$
The numbers I give above are for $d\rho/dT$.  Temperature increases and the effective price you pay for your transportation fuel increases.
A: On the other hand most of the car's energy goes to overcome air resistance (at least at highway speeds) and drag is proportional to air density.
And air density is  inversely proportional to  absolute temperature. So on hot summer days you might not have to burn quite so much - assuming you don't need air conditioning.
