Why is this MOSFET Differential Pair On on the Right-Hand Side?
Your circuit model uses an ideal current source. This source will lower its sinking terminal to as low a voltage as necessary to obtain 25 uA current. In your circuit that means lowering it to -0.5 V and turning on M1.

If you implemented this circuit on a chip, you'd use a current mirror there, not an ideal current source, so the behavior will be different.

If you want to get at least slightly more realistic behavior, you can include a resistor in parallel with the current source. But you'll still end up with a negative voltage at Vs, which isn't what will happen in the real circuit.

If you more realistic behavior than that, you will probably need to design the actual current mirror circuit and include it in your model, so that the compliance voltage behavior of the source can be included.