In (1) you have a source follower and this means that the MOSFET provides no extra gain that might cause the op-amp to become unstable when negative feedback is applied.
In (2) you have a source follower and this means that the MOSFET provides no extra gain that might cause the op-amp to become unstable when negative feedback is applied.
In (3) you have a common source MOSFET that will have shed-loads of voltage gain and will inevitably cause oscillation when the feedback loop is closed.
Why does positive feedback from the high side of an nMOS cause
Actually it's negative feedback at DC and low to middling frequencies but, inevitably the extra gain is going to tip the balance good and proper. The LED has a dynamic resistance of a few ohms and the drain has 70 ohms; maybe an added voltage gain of 25 times. Do you think op-amps are designed with this much spare margin to cope with a gain of maybe 25 times within the feedback loop?
Extra questions answered in comments
I don't understand why the gain is dependent on the ratio of source
and drain resistance
M3 has gain due to the very small dynamic source resistance (circa 3 ohm from the LED) and, the 70 ohm in the drain. Given that source and drain currents are virtually the same, a signal voltage on the source will be magnified on the drain by 70/3 = ~25. Then, because the source signal voltage is roughly what the gate signal voltage is, you get a gain of circa 25 from gate to drain. And paraphrased later: -
For both a BJT and a MOSFET, the collector/drain current pretty much equals the emitter/source current and, because the signal at the gate/base is pretty much the same as that seen at the source/emitter, the magnitude of the signal at the collector/drain is higher by the ratio of drain resistor to source resistor. Gain approximates to \$R_D/R_S\$ and with an LED in the source the dynamic resistance will be a few ohms.
Whether it's a quirk of the discrete nature of the sim itself or
there's some source of noise in the model that I don't see, or
something else entirely-- even if it's a bad design with gain in the
loop, the oscillation still needs to start somewhere, no?
Even if it's unstable, surely it needs some initial perturbation to
start the oscillation? Like an inverted pendulum or a ball on top of a
hill -- it can hold its position forever in a motionless vacuum, even
if the slightest nudge will cause it to fall.
The very act of a simulation circuit being activated will cause power supply transients to initiate oscillation hence, we can use simulators for designing oscillators (and I have done so many times). There is one pre-condition though; you must prevent the simulator trying to calculate DC conditions first because, that can stop the transient condition triggering oscillation. You may have heard that simulators don't work with oscillators but, that is largely untrue. Simulators are pretty decent these days providing you use them properly.
Is there some feature of the simulation algorithm that intentionally
adds perturbations to check for metastability?
Metastability is a digital phenomenon and isn't related to this question.
A simulator (when used appropriately) instantly applies voltages to the circuit. This transient is enough to begin the process of oscillation.
I'd like to also understand what is underlying not just the
oscillation but why it's not getting to the rails. Is it reasonable to
conceptualize this as analogous to a PID loop? If the Kp is way too
high, I'd expect it to hit the rails; is there some component doing
derivative operation to limit the swing?
Because the gain is on the cusp of "enough" and, usually, in so-called linear circuits, they behave not so linearly closer to the power rails and, the gain reduces. Nothing to do with PID controllers. No derivate just soft clipping leading to hard clipping.