With a color code of Brown, White, Green, Red, Silver the resistor is actually 19K5 at 10%
A 19.5KΩ doesn't seem to make any sense there to me. Something like 1Ω does, though.
In the circuit, it is in parallel with a 12V TVS diode (SA11). 1 ohm doesn't seem like it should be across a diode. This diode has cathode connected to Source of main switching transistor, and anode connected to one of the primary windings and the negative side of the input capacitor (150uF 450V).
Is there a reason a 1 ohm resistor would be there? Any why is it colour coded like that?
Yeah, it actually does make sense, since it is the current sense resistor. (There's a tongue twister!)
It appears to be the current sense resistor between the MOSFET source and the 0 VDC (side of the +370 VDC input,) with a 12V TVS across it to prevent any spikes from exceeding 12V heading back towards the current sense input on the SMPS controller IC:

BTW, sorry, I meant to have flipped the top side to make it easier to compare to the bottom of the real board, I accidentally flipped the wrong image when I was pasting them together....
From looking at other DIP-8 SMPS ICs, and the tracks for this one, I think pins 5 and 7 are the power pins.
The multimeter shows it oscillating between about 11.5 and 13.8V at about 0.5Hz. That sounds more correct.
Yeah, it seems to have the same pinout as some other common 8-pin SMPS controller ICs like the UC3842/3/4/5. Those particular ones happen to have a minimum startup voltage of about 16V, and lock out below 10V.
C8 measures 1.2nF out of circuit. I don't have the tools to measure leakage. Would you expect 1.2nF for supply filter for a SMPS IC? Seems low compared to logic ICs.
...
I can probe what you've likely deduced as the power supply pins of the UC80669 with a scope. I would need to use my isolated step-down transformer.
Yeah, C8 is the timing capacitor, it and a resistor sets the operating frequency via an RC constant.
I take it you already double checked that you didn't get a capacitor in backwards or something?

I would take a close look at the power coming from the auxiliary winding, make sure it is actually starting to generate power into the auxiliary supply reservoir cap since it looks like the controller IC is only bootstrapped from the +370 DC rail, then is supposed to be powered from that auxiliary winding on the primary side. The controller IC obviously needs to maintain voltage on Vcc to continue operating after trickle-charging up from the +370VDC.
You could check to see (I would even scope it, multimeters are slow) what happens with the voltage across that resistor, see if it looks like it is actually sensing high current and going into overload due to the current sense input "seeing" too much current, (which would point to some sort of potential overload or short on the output side,) although you
do have LEDs come up, so it can't be a
total short.
I still suspect it is probably a primary side power drop-out to the controller IC, but it still could certainly be lots of different possible things going awry, I suppose.
Edit: If it were me, I think I would put any removed components back in, except pull the MUR615CT out, tack a 5W resistor of something like 470Ω or 1KΩ across the secondary winding so it doesn't try to generate crazy high voltage spikes on the secondary winding, then power it up and see if it runs. If you measure the AC voltage across that temporary secondary resistor (yes, it will be high frequency but even most multimeters will show something resembling a measurement) you should see something, probably something like 25-30 or more volts AC because it should just run full-on in open loop.
If it runs, then your problem is probably something on the secondary side.
If it
doesn't run open loop like that, then there must be a primary side problem that needs to be addressed first before it will ever work driving the real laptop output circuitry on the secondary. If it doesn't run you'll have to check things like the power to the optocoupler collector, the controller IC, etc. that are on the primary side that might be dropping out.