How induction heating works
I am considering your edited case, where the frequency and current are held constant. You observe that the EMF should be the same in both pots, therefore the pot with a lower resistance should heat up more. However, I believe you are assuming something that is not true. The EMF will not be the same in the steel and aluminum pots because the magnetic flux density (B-field) will not be the same.

The primary coil, driven at constant current, will give rise to a time varying magnetic field (H-field) which will be the same in both pots. But due to the high magnetic permeability of the steel pot, the magnetic flux density (B-field) in it will be much, much higher than in the aluminum pot.

As you know, Faraday's law relates the rate of change in the B-field to EMF. Since the steel pot will have much higher B-field, it will also have a much higher EMF, and thus a much higher power dissipation despite its high resistance.

If the EMF in the two pots really were the same, then you can easily see from the equation P=V^2/R that the pot with lower resistance would have higher dissipation.

To be just a bit more numerical, note that the permeability of steel is around 100x higher than aluminum, so the EMF in steel is actually around 100x higher, and V^2 is 10,000x higher. Aluminum is a better conductor than steel, but not 10,000x better, so the net result is that the steel gets much hotter under the same H-field magnitude.