Step down from USB (5 V) to 4.3 V for battery dummy
Assuming that the powerbank voltage has a regulated output at 5.1V, then a series diode would do it, probably giving you 3.9-4.1V under load. 
Using a bipolar transistor as a diode (join B+C) is less spongy giving you closer to 0.7V drop under load.

Just check the no-load voltage in case your load goes to really zero, and leakage reduces the diode drop.

Alternatively just open your powerbank and either change the regulator feedback to get the voltage you want, or if it is a single cell power bank, just bring the battery out directly.