So far as I understand it, all of the current that flows through the capacitor has to be burned off (by the load, Zener, in-line resistor, rectifier, etc.) so that is energy lost.
This is a particularly bad circuit so far as efficiency is concerned. The smoke detector will need--in the worst case--the current to drive the Piezo element in the event of a fire. It will likely need around 10mA to sample for smoke every 10 seconds, or so. All other times it is likely burning no more than 1mA. They likely draw 50mA of current from the battery during a low battery test but--unless their circuit is incorrectly designed--that should come exclusively from the battery and never from mains.