You start your question with a false assumption: that something like a 60 W light bulb has a constant power. Instead, it has a constant resistance, and is rated for 60 W wherever it is intended to be used.
A resistive load (like an incandescent light bulb) rated for 60W is assigned that rating based on the intended supply voltage. Using Ohm's law we can determine some other properties. We know the intended voltage (120 V) and its power rating (60 W), so we determine its resistance:
\$R = E^2/P\$
\$120^2 / 60 = 240 Ω\$
Now that we know the load (bulb) resistance, we can apply the higher voltage (240 V) to see what happens.
\$P = E^2/R\$
\$240^2 / 240Ω = 240 W\$
Since the filament in the bulb is not designed for 240 watts of power dissipation, it will burn itself out quickly.
Other loads, like computer power supplies, can sometimes operate at different supply voltages. (Sometimes it is necessary to toggle a switch, and sometimes the supply can auto-compensate.) In switching power supplies like that, they can often accommodate a wide variety of supply voltages. The device depending on that power supply can then consume a more-or-less constant power from one region to another.
A device that can handle different voltages doesn't alter its resistance to maintain a given power output, and therefore is not said to be a "resistive" type of load. Instead, it could be characterized as an inductive and/or capacitive load.