Trouble Calculating Transistor Voltage Drops
You only have 2 states here, assuming that battery voltage is at 4.5 volts.

In state 1 the switch is OFF and the CE voltage rises to the LED turn-ON voltage, about 3 volts. R1 prevents the LED and Q1 current from becoming dangerous. Now the LED is ON and the BE voltage should be much less than 700 mV, which is internal leakage in Q1 due to an absence of a resistor from base to emitter.

In state 2 the switch is ON, the BE voltage is now ~700 mV. The CE voltage drops to just hundreds of mV, based on the ON resistance of Q1. The low CE voltage causes the LED to go out, as its turn-on voltage is about ~2.9 volts.

Assuming the battery is plugged in these are the only 2 valid states for this circuit.