My Toshiba HX-10 gets mighty hot!

The B435 can be replaced simply be a BD242. It support a very high temperature. The transition frequency is much less important but I do not think it is important for this use.

I think it should heat a little less but I'm not sure. The main thing is that it supports the power to be supplied. Do you have a thermal probe? If it gets too hot, think about changing the nearby capacitors. If the temperature exceeds 80°, capacitors at 105° are required.

I ordered a z80 cmos at $1.16 to try. This solution is probably sufficient and it is the cheapest.

A transistor just burns off the excess voltage, like a 78xx or any other linear regulator would. Efficiency depends on the ratio out:in voltage, which here is 5V/9V = ~56%. For example @ 1A that is 9W in, 5W out = 4W dumped in the transistor. Replacing transistor with another type changes nothing to that. Zero. Nada.

Max. junction temperature is rated as 150 oC for both 2SB435 and BD242, which is pretty much standard for transistors like this. Btw the BD242 has a much lower DC current gain, so isn't a good replacement anyway.

For a switching DC/DC converter: say it's ~85% efficient. Then at 1A it's 5.9W in, 5W out, 0.9W lost as heat in the DC/DC module. @9V in, that's 0.65A pulled from transformer vs 1A with transistor or 7805 regulator. And 9W - 5.9W = 3.1W less heat released inside the machine.

Thank you RetroTechie for clear advice.
I only said that the BD242 would heat a little less because as the diodes more the intensity exceeds the nominal value, more the heat increases. If the nominal value is never reached replace the transistor will indeed not change anything.

The z80 CMOS, the heat sink replacement and the addition of two heat sinks for SDRAM should suffice. We just have to find the right heat sink and an effective thermal paste.
About capacitors, I checked the two closest (2200μF 35V and 4700μF 16V). These support a temperature until 85°. We must check if the temperature exceeds 80° if so, replace them with new capacitors at 105°.

(PS: The ratio out:in voltage, which here is 5V/9V = 44.44%)

You are sure that 9V - 44.44% is not 5V?!