Interesting thread... Remember, it's current, not power, that heats up electrical components. Let's see if this makes sense...
P=IV ; Power (W) = Current (Amps) * Voltage (Volts)
V=IZ ; (Z=Impedance, ohms)
P=I(^2)Z
So... halving your impedance (4 to 2 ohms) at the same power output will quadruple your amp's current draw, and thus also quadruple resistive heating of its components! *this is assuming a 100% efficient power amp...
BUT...
SPL~P*Efficiency
Two cabs generally increase efficiency by 3db, right? Which is equal to approximately a doubling in electrical power. So you'll turn your amp down by 33% to get back to your original SPL, reducing resistive losses some, but not all.
BUT BUT BUT...
It doesn't really work that way, because amps aren't 100% efficient (or else they wouldn't need cooling at all...duh...). So the less efficient your amp is, the more dramatic the increased resistive heating due to lower impedence cab.