But with your point about the coating, I'm only off by a factor of ten...
1000 times smaller beam diameter means 10^6 times higher irradiance. With 1% absorbance it's still a factor 10^4 off.
how much power do we really need on target to make it overheat given that it's designed to cycle between daylight and night
The solar panels are designed to receive 1500 W/m2 irradiance from the sun, of which maybe 500 W/m2 comes out as electricity and the other 1 kW/m2 is heat load. A black-body radiator against 4 K background temperature on one side and earth (279 K) on the other side will reach an equilibrium temperature of 330 K (57 C). To heat the solar panel to a damaging 150 C (just a guess), you'd need about 2.5 kW/m2 extra (on top of the solar irradiance).
It wouldn't surprise me if the back side of the solar panels is used as a radiative heat exchanger to get rid of the heat generated inside the body of the satellite. The gold-coated foil around the satellite body acts as a kind of thermos bottle to insulate the electronics inside from extreme temperature swings, but still the electronics need to get rid of its own dissipation, so you would not only destroy the solar panels but indirectly also the electronics.
I doubt that it's practical to aim specifically for the solar panels for an object that's travelling at 7 km/s, with a mechanism that can do that within 1 microrad. It might be more practical to have a 10 m diameter spot size, which would require 250 kW, which might be doable with a CO2 laser at 10.6 micrometer wavelength (the atmosphere seems to be fairly transparent at that wavelength).
If the solar panels have a mass of 15 kg/m2 (just a guess), then you would need to maintain this 2.5 kW/m2 for about 10 minutes, over which time the satellite will be travelling 4200 km. This doesn't sound easy...