Any voltage below the logic LO input threshold voltage will be recognized as a LO by the demux. As you increase demux supply voltage, this threshold just keeps getting higher. So any LO signal output by the MCU will still be below the threshold and that means your MCU can still send zeroes to the demux.
BUt the HI voltage threshold for the demux increases too, and your MCU cannot output a HI signal to exceed this voltage threshold. Even when the MCU tries to send HI signal to demux, the voltage is so low that it is less than the logic LO threshold so your demux will see it as a LO signal.
So your MCU can still send LO to the demux, but not HI. So if switching channel 0 only requires all LO signals, your MCU can do it. But it can't do anything that requires a HI signal.
You could use a pull-up resistor to 12V on the input pins of the demux and use a pull-down logic-level MOSFET that has it's gate driven by the MCU. SO when the transistor was off, but the resistor would apply a HI to the demux input, but when the transistor is on it would pull the line LO.
Do you understand what I mean when I say pull-up resistor and pull-down transistor?
I'm curious...how did you get it to work on the breadboard? Was the demux also being powered by 12V on the breadboard? Or by the same voltage as the MCU?