Electronic – Why is the voltage drop across this transistor the full voltage of the battery

I see you have several answers, but they are mostly off the mark. You are NOT seeing a drop on the LED due to leakage of the transistor. Adding a resistor to ground on the base won't fix anything since you essentially already have that between R1 and R2. These transistors do have a small amount of leakage with the base held at 0V, but again, that's now what you are seeing.

What you are seeing is the voltmeter acting like a resistor, which pulls down on the LED cathode enough to get the little bit of current it requires. LEDs are diodes, so the current at a function of voltage is quite nonlinear. With the voltmeter providing a small current path to ground, the LED forward voltage is apparently 1.55V. It is probably at 2V or so when really on. 1.55V is plausible to support the few µA or even nA that the voltmeter draws.

To prove this point, use the same voltmeter to measure between the LED cathode and the 5V supply. If the circuit causing the apparent voltage drop on the LED, you should now read about 1.5V. I predict you will read essentially 0V. That is because now there is no current path thru the LED to ground.

The LED cathode is a very high impedance node with the switch is open. So high that the voltmeter significantly effects the circuit.

You need a high-side switch to get this to work, as in this answer

Q3 shifts the 3.3V level to your 4.7V (rather than the 9V shown) and R2 supplies base current for Q2. R3 prevents leakage in Q3 from being amplified by Q2.

Pick R2 to give about 1/10 to 1/20 of the load current. R3 can be a few times larger than R2 and you can make R1 of similar value to R3.

Q2 should saturate when on and give you tens to hundreds of mV drop from the 4.7V rail depending on the transistor and the load current. If you substitute a logic-level p-channel MOSFET for Q2, you could make the drop even lower under favorable conditions. For example, a 5m\$\Omega\$ Rds(on) power MOSFET would have a drop of only 50\$\mu\$V at 10mA.