Flux density in a transformer after switch off
There is a mechanism for resetting the flux in the core eventually, and that's the mechanism you explicitly chose to ignore in your question, the copper winding resistance.

With a truly lossless inductor, at a certain flux (which means at a certain current), with zero volts on the input, the flux and current will remain at those values indefinitely. Consider what happens in a superconducting magnet, when shorted by its internal superconducting switch.

I'll wager that your 1800VA toroidal is wound with copper rather than superconductor. What happens is the current through the winding resistance generates a voltage, which reduces the current, with a time constant of L/R. Plug that into the normal exponential relation to find flux and current as a function of t. Of course L/R is infinity for the superconducting case.

A dropped half-cycle is just equivalent to more volt.seconds than usual, so would be expected to saturate the inductor.

It's interesting how the recovery from a switch-on saturation occurs. Although the average DC voltage applied from the supply is zero, and so you would not expect the flux to slew cumulatively in either direction, in saturation the transformer draws a heavy current, creating a larger voltage drop in the supply and winding resistance. This provides the net average voltage which gradually resets the flux to mean zero. With a lossless transformer, and stiff supply, the transformer would continue to saturate for a period every cycle. The same argument can be used to explain why a unidirectional (diode connected) load on the secondary of a transformer can drive it into saturation.