Where to start?
First, your HV unit is not a transformer. It is a high-voltage module. It puts out a pulsing high voltage.
So, the high voltage transformer was made to make arcs, and as such
does not allow ... a no-load situation.
You are correct. Your capacitor will store charge and will destroy your module. Furthermore, you have taken the input power specification, (and it would be nice if you would share with us where you got the numbers) of 18 watts at 4.5 volts, and then calculated an output power of 18 watts/20 kv, giving .9 mA. This is wrong. Input power does not equal output power.
You have also failed to notice the part of the description that reads
Work: Input voltage 1.5V ~ 3V can work for about 1 minute, the input
voltage exceeds 3V continuous work does not allow more than 30 seconds
In other words, you will not be able to run your unit for more than 30 seconds at a time. And it doesn't say how long you need to let the unit recover from 30 seconds operation, either.
Having failed to understand your HV unit, you have equally failed to understand your water capacitor. To begin with, ultrapure water is does not have an infinite resistivity. It is, in fact, 18 Mohm-cm. For your described capacitor, this amounts to ~ 2 Mohm. At 20 kV, that will require 10 mA which you cannot provide. You mention a "coating on the plates" which you believe will prevent current flow, but you do not describe it, and I hope you will forgive my suspicion that it may not work as you plan.
You speak of your desire not to break down the dielectric (the water) and then talk about how pulsing or AC voltages will avoid this. You seem to be unaware that the breakdown potential for pure water is ~3 MV/m. Since the spacing for your capacitor is on the order of 12.5 mm, the breakdown voltage will be ~ 37 kV, or twice your voltage, and you have no need to worry.
Given the dielectric constant of water (80), the effective capacitance of your capacitor will be ~ 100 pF. Since your module has a diode in its output, you are correct in thinking that you will get just a DC output, since there is no major discharge path. You cannot discharge your capacitor from the input side. Any such discharge mechanism would have to be synchronized with the HV pulses, and would have to withstand 20 kV. Such a circuit will be neither simple nor cheap.
Your selection of series resistor is very strange. Your calculation for voltage drop make it clear that you have used the input figure of 4.5 amps, and this has absolutely no application to the output.
So, overall, I have no way give you advice, except to suggest that you do something else. I don't see any instance where you seem to actually understand how your proposed circuit would work, or why, and that is not a good starting point.
It's quite possible there is an RC snubber or MOV across the switching device which passes enough current that your meter shows 120VAC. I can't quite make it out from your haywire setup, but it looks to be a relay with spade terminals for the power. It could also be a triac, likely mounted on a heat sink.
In such a case, you won't see the switch functioning unless there is a credible load in place of the transformer. Something like a light bulb would work.
Be extremely careful, even with the lethal high voltage, and potentially dangerous magnetron excised, you've got potentially lethal exposed mains voltage.
Best Answer
I believe the way microwave oven magnetrons are wired is that there are two coils on the transformer more or less in series, one to supply the HV and one to supply filament. The filament doubles as the cathode and it has some series inductors to keep the RF from leaking out the power terminals. The HV return from the anode could just be the metal case as it is less than an amp. An inverter-driven microwave likely has the same setup, however the transformer will be smaller as the switching frequency will be much higher (60 Hz vs several kHz). The inverter should have essentially the same connection to the magnetron as the older style single transformer solution. Also, you won't be able to run the inverter transformer by itself, you'll also need the rest of the inverter drive and control electronics. I'm not sure if it will work without the front panel/user interface; it may be required to turn the inverter on and possibly select the power level in some way.
However, turning on a magnetron outside of a case is very dangerous. Not only could you kill yourself with the high voltage, the high power microwaves could give you serious burns as well as possibly causing interference and damage to electronic deivces. The magnetron could also overheat due to reflected power if it is not properly coupled to a waveguide.
Edit: Looks like the inverter microwaves operate on exactly the same principle as the older ones, just at a much higher frequency. You'll either need to keep the front panel or figure out what signal it sends to the inverter to turn it on.
Image from http://www.electronicspoint.com/threads/microwave-inverter.234684/