The DC across the switch will not be the full B+ voltage. It will only be the difference in B+ with the switch open or connected.
With a reasonable sag resistor value, that will probably be no more than 50V.
Or am I missing something?
Sag resistor is not a factor. When the STBY switch is closed there will be zero volts across it. When the STBY switch is open there will be full B+ across it, ie, B+ on the rectifier side, zero on the filter cap side.
As I understand switch DC ratings, they define the maximum voltage and current across the switch when you flip it. Lower current typically allows higher voltage. Unfortunately, most switch data sheets only list DC ratings for high currents such as 10A. We don't really know what they can withstand when we're in the milliamp range.
I'm assuming the sag resistor and switch would be between the rectifier and the and main filter caps. The sag switch would short out the sag resistor when closed.
If the standby switch is open, there is only roughly 1mA across the sag switch (it's limited by the 220k+220k filter cap bleed resistors). If we assume a 200R sag resistor, that's 0.2V.
If standby is closed, the voltage across the switch is limited by I*R with regard to the sag resistor. Not sure what the typical current draw of a Twin is, but let's say 250mA, and we get a voltage across the switch of 50V.
The voltage across the switch would go up momentarily with the inrush current at turn-on, but for a very short amount of time. This would only be a problem if you flipped the sag switch at the exact moment of the inrush current.
The insulating rating of the sag switch would of course have to cope with full B+.
Please correct me if I'm wrong!