Drawing only the PMOST bit of your schematic:
EDIT2: I completely forgot to add the 10k pull-up resistor. It's added now, R4. Sorry.
simulate this circuit – Schematic created using CircuitLab
Basically when the mosfet is turned on the R3 in my schematic puts a companion current into the transistor and the zener diode will have to conduct a little less current to shut off the transistor, reducing the shut-off voltage a tiny bit. When the MOST shuts off the support disappears and the circuit will have a much better chance to stay stable. Now, beware that R2 and R1 have to be able to pull the base of Q1 (in my schematic) below the 0.6V threshold, so R3 should not be too small and R1 and R2 not be too big. If it doesn't turn off at all I mis-estimated the transistor and you may have to increase R3, although I think it's close enough.
EDIT1: In response to your simulation: As I stated, I could be off with the transistor. What you see is the voltage drop across the diode, so maybe the system just doesn't shut off, or much too late.
Try simulating with a larger range of voltages to see what happens, then see if a larger R3 helps lifting the shut-off voltage. Other than that you can tweak the zener, though that will also tweak the LED trigger. Do keep in mind simulations aren't holy. I could model the whole transistor and do a full range calculation of the currents and cut-offs but that's something I haven't done in a very long time, so I'll probably only have time to really sit down for it in quite a few more days.
It's also possible a lower frequency may show different results, but that's quite unlikely, since from the perspective of discrete transistors and mosfets 120Hz is already pretty much DC.
EDIT3: When I wanted to do a calculation I looked at the schematic and saw I forgot the pull-up resistor in my schematic, R4. It's an absolute oversight on my part, but these things do happen. Now that it's added, I have run my own circuit lab simulation with the following schematic:
simulate this circuit
With these results:
You can see it switches off at about 5.5V, and back on at about 6.3V: Hysteresis! As said, you may have to change the zener a little, if the 9V battery is rechargeable, it may be pushing it a little, although that's still 0.92V per cell, so they should survive it well enough.
As you increase or decrease the R3, let's say between 90k and 330k the hysteresis window will change, with 90k being a quite large window and 330k being a tiny window.
What I think you're seeing is the start of an exponential where the capacitor is charging. The time constant is 22s and the pulse duration is very small compared with this, so you're only seeing the early stages of the exponential.
In charging a capacitor there is a flow of power in, but the resultant energy is stored and not dissipated.
I don't know why there is no discharge curve; it may be worth decreasing the pulse frequency significantly so that we can see what's happening in the time frame of the time constant. The integration sampling time may also be a factor.
Best Answer
Actually it's the parasitic drain source capacitance that is inside the MOSFET that is producing the pulse at the output. Nothing really to do with conventional MOSFET action.
The FDC5614P has an internal capacitance of 90 pF and the CR time with R3 is 90 ns. At 5x CR the transient waveform should be pretty much over so that is a time of about 0.5 us and this looks about right on your graph given that you have a scale granularity of 2 us and my powers of interpolation aint that good!
Try shorting gate and source and removing C1 and R2 and repeat the experiment. I bet it looks very similar.