Output current of step-down switching regulator
If you're using a 90% efficient DC/DC system (IC, inductor, diode, etc), then output will be the 277mA you calculated *0.9 (250mA).

The "charges" come from the input. Nothing is "gained". You exchange voltage potential for current (and heat, maybe some noise, etc).

1 - Power = Current * Voltage

2 - Power in = P out + P losses

Pin = 5V * 100mA = 0.5W

Plosses = Pin * inneficiency = 0.5W*0.1 = 0.05W

Pout = Pin - Plosses (or Pin * efficiency)= 0.5W - 0.05W (0.5W*0.9)= 0.45W

So, if Vout = 1.8V, Iout = 0.45W/1.8V = 0.25A

p.s.: In step-up systems the reverse is true. The output will be capable of less current throughput. The "original" current does not go anywhere, you just exchange it for more voltage potential (and heat, etc).