Voltage drop of a PWM signal
With the output shorted and 100% PWM a current of ~5 V / 200 Ω = 25 mA is expected. At 50% PWM it should be 25 / 2 = 12.5 mA on average. Your circuit draws 6.22 mA with no load, so this must be added to the load current to get total current. 12.5 + 6.22 = 18.72 mA. So far so good.
The only problem is you are not getting zero volts across your 'short', but around 0.5 V. Assuming 0.5 is the average voltage, the 'shorting' wire must be dropping ~1 V during PWM 'on' time (assuming 50% PWM). This should reduce the load current to ~4 V / 200 Ω = 20 mA peak, or 10 mA average. So you have an apparent anomaly of ~2.5 mA.
However this small 'anomaly' may vanish when more accurate measurements are made. Depending on PWM frequency, a DC meter may not measure the DC component of the PWM waveform accurately. At high PWM frequency the transistor may turn off more slowly than it turns on, increasing the effective PWM ratio. If the supply leads have significant inductance and/or the power supply has poor stability then the supply voltage may increase under load, and the flyback diode could produce a voltage offset of up to ~0.8V. Another possibility is a bad connection at the 'short', producing randomly changing load current and voltage across the 'short'.
If you use the same meter for different measurements then the current and voltage could change from one measurement to the next, and you wouldn't know. To see what is really happening (including inductive effects etc.) you should look at the waveforms with an oscilloscope.