Only if the MOSFET can handle the required voltage and current. Typical solid state relays will have protections built in against short circuit, EMI, electrostatic discharge, over voltage on the signal or power supply, and provide sufficient heat sink.
MOSFETs inputs do have a small capacitive load component, normally benefiting from a MOSFET driver in use with other semiconductors, but metallic switches, particularly to incandescent bulbs, already have to deal with the surge current at turn on.
What is non-standard about on and off? Besides when my bi-metallic blinker in my car stops working, I can be sure I have a light bulb that I need to replace. I don't know that with the electronic flashers.
- bimetallic blinker with load and my board = failure.
- bimetallic blinker with load without my board = success.
- electronic blinker with load and my board = success.
Something is keeping the bimetallic blinker from doing its thing. It outputs a constant 7 - 7.4 V and does not light up the load even dimly. Which tells me the MOSFET is kept off and that must be because MCU is not turning it on.
Is there such thing as minimum current through a voltage regulator ? It is a MCP1799 with a max spec of 80 mA and no min spec. Its output voltage is 3.3V as expected but maybe, due to how the bimetallic starts with a very low current, maybe it is just not enough to power the MCU ?
Right - there must be a heating element in the flasher module that requires enough current to cause it to function.
The usual method is to add a resistor to simulate the bulb resistance to draw enough current to heat the heating element; that resistor may need to withstand 5 or more Watts and will get rather hot.
There are also "heavy duty" flasher modules that do this internally. The reason is that as the electrical load increases the typical duty flashers turn off faster.