Electronic – the ground-source voltage of this current mirror

Referencing the diagram from the linked article:

The drain current of either transistor is mostly determined by the respective gate-source voltage.

For the lower transistor, the gate-source voltage is just the input voltage

$$v_{GS1} = v_{IN}$$

For the upper transistor thought, the gate-source voltage is

$$v_{GS2} = 0 - v_{D1}$$

where, in this example \$v_{D1} < 0 \$.

Now, the drain voltage of the lower transistor \$v_{D1}\$ will, essentially, be whatever it needs to be such that the upper transistor transistor current matches the lower transistor current.

For example, if the input voltage rises, the drain current of the lower transistor will increase which will act to reduce \$v_{D1}\$ (make \$v_{D1}\$ more negative) which increases \$v_{GS2}\$ thus increasing the drain current of the upper transistor.

The current can still flow through the "substrate" even though the channel is pinched. The reason why it saturates is that there will be a region of higher resistance of size proportional to the Drain-Source voltage, and therefore the resistance of this region will be proportional to the same voltage.

But as current is voltage/resistance, the dependence will cancel out and you'll get "constant" current.

From Wiki (emphasis mine):

Even though the conductive channel formed by gate-to-source voltage no longer connects source to drain during saturation mode, carriers are not blocked from flowing. Considering again an n-channel enhancement-mode device, a depletion region exists in the p-type body, surrounding the conductive channel and drain and source regions. The electrons which comprise the channel are free to move out of the channel through the depletion region if attracted to the drain by drain-to-source voltage. The depletion region is free of carriers and has a resistance similar to silicon. Any increase of the drain-to-source voltage will increase the distance from drain to the pinch-off point, increasing the resistance of the depletion region in proportion to the drain-to-source voltage applied. This proportional change causes the drain-to-source current to remain relatively fixed, independent of changes to the drain-to-source voltage, quite unlike its ohmic behavior in the linear mode of operation. Thus, in saturation mode, the FET behaves as a constant-current source rather than as a resistor, and can effectively be used as a voltage amplifier. In this case, the gate-to-source voltage determines the level of constant current through the channel.

Also, from the MOSFET operation description, under saturation:

Since the drain voltage is higher than the source voltage, the electrons spread out, and conduction is not through a narrow channel but through a broader, two- or three-dimensional current distribution extending away from the interface and deeper in the substrate. The onset of this region is also known as pinch-off to indicate the lack of channel region near the drain. Although the channel does not extend the full length of the device, the electric field between the drain and the channel is very high, and conduction continues.