Electronic – Use transistors for controlling two LEDs

It seems that the grey wire has a high output resistance so the more current you source the more voltage drop you get, although I can't explain why it is able to provide about 13mA to LED1 with no drop and then drops to 2.5v with a few more mA.

Anyway, the circuit I suggest used two transistors, a PNP and a NPN each driving one led.
The benefit of the circuit is that it only draws a couple of mA (depends on the gain of the transistors) from the gray line so you will get no drop there, the main current for the leds is supplied from the purple supply line.

There is no need for a base resistor because the base-emitter current is limited by the resistors connected to the emitter.

I found this ATX12V Power Supply Design Guide linked from the wikipedia page for Power Good Signal and it describes the specs of PWR_OK (grey wire)

Based on these specs the current you can source is negligible so I don't know how you manage to get 5v while sourcing more than 10mA (as shown in your circuit with only led1 connected). I have also seen this wiring scheme from the page you linked which also connects the anode of the led to the PWR_OK pin and sources several mA.
I assume that the specs about sourcing are the minimum and designers actually use output circuit with more current sourcing ability.

If I was to base a circuit on these specs then I would use a pull-up resistor (R4 below) to drive the base of the NPN or even better a mosfet.
The PNP circuitry doesn't need to be changed, sinking 1mA will be enough and is within specs.

This is the simplest circuit I can think of (2 parts):-

^{simulate this circuit – Schematic created using CircuitLab}

Input = 5V Red Input = 0 Green

The circuit works because the \$V_F\$ of a Red LED is generally much less than the \$V_F\$ of a Green or Blue LED.

Here is a typical RGB LED datasheet:

And here are the typical plots of \$V_F\$ vs. current:

When the transistor is "off", the Green LED will have a forward voltage of about 3.0V at 10mA. When the transistor is "on", the Red LED will have a forward voltage of less than 1.9V. Referring back to the Blue/Green graph, little current will flow through the Green LED at that forward voltage.

This particular arrangement shows a common-cathode arrangement, however it could be flipped and a similar P-channel MOSFET used. In that case, 0V would select the Red LED, and +V in would select the Green LED.