Electronic – ON/OFF safe implementation

Why not use a relay - it can be driven from the 5V logic, possibly with a transistor to help it and the relay contacts would just connect in parallel with the load. What you haven't said is how fast you need to do the load bypass switching (is there any PWM involved) and what the maximum load current is - this may drive you down the MOSFET route.

Here's the relay I'd consider if it is just to act as a dumb load on/shunt switch: -

BTW - in your 2nd circuit there is no gate over-voltage protection when the power supply is greater than 20V (most MOSFETs can deal with maybe 15 to 20V on their gate but not 50V).

This would be a very simple modification that might be acceptable.

^{simulate this circuit – Schematic created using CircuitLab}

When the switch is closed, \$Q_1\$ is off and the red LED is also off. When the switch is open, \$Q_1\$ receives base drive via the green LED and \$R_2\$ and \$R_3\$, sufficient to turn on the red LED. The green LED current will be cut down by a factor of about 20 (about \$1\:\textrm{mA}\$), so it won't be nearly so visible. There may also be some "leakage" into your pump/PIR circuit via \$R_3\$, but I don't think that will cause you trouble. You might want to lower \$R_1\$'s value, just in case. So this may, or may not, be acceptable in your circumstance. You'll have to decide. But it's easy to understand and implement.

Just for completeness, the "inverter" circuit that I believe was suggested might look like the following:

^{simulate this circuit}

But the price paid here is that the red LED current either goes through the red LED or else it goes through \$Q_1\$. But the current is always present. So you pay an added price in current draw. That may also be acceptable, though.