Given the same amount and temperature of tap water, wouldn’t an 800W water kettle take twice the time but the same amount of power to boil the water as does a 1600W kettle
To answer your question directly, yes... and at the same time, no.
Yes: The same amount of energy is required to heat a fixed quantity of water from one particular temperature to a second higher particular temperature no matter whether the energy is applied at an 800W/second rate or a 1600W/second rate, so the 800W kettle requires the same amount of power as a 1600W kettle, at 1/2 the rate, expended over twice the time. Theoretically, in an ideal world.
No: In the case of an electric water kettle drawing power from a lead acid battery storage system, given the same physical wiring, battery storage, etc., a higher current draw results in more losses in the system delivering the power to the kettle that heats the water than a lower current draw. The biggest practical loss may be the one mentioned by john61ct,
Peukert's law. Simply put, the usable energy storage capacity of lead acid batteries depends on the discharge rate; higher discharge rates lower the usable energy storage capacity. So using the 800W kettle to heat the water reduces the usable energy stored in a lead acid battery system by less than using the 1600W kettle (but takes twice as long). Practically, in the real world.
Assuming, of course, that the rate of loss of heat energy from the system during the heating period isn't high enough to make a significant difference over the increased time required. I.e., that both the 800W kettle and 1600W kettle have decent insulative qualities and/or that the quantity of water is small enough that the desired increase in temperature can be achieved in a relative short period of time.