Once the electrons reach the end of the channel, they experience the high electric field in the depletion region surrounding the drain junction and are rapidly swept to the drain terminal. Therefore, the device still conducts, but I can't understand why the current is unchanged.
Example N-JFET: The width and the form of the depletion region is determined by VGS (general width) as well as VDS (unsymmetric form). The channel width becomes smaller in the drain region (equivalent to a broader depletion region).
Small voltages VDS (max 1 V): The channel behaves (nearly) like an ohmic region because the influence of VDS on the form of the depletion region is not yet remarkable. Hence, ID increases (nearly) linearly with VDS.
Rising VDS: The channel width becomes smaller close to the drain region and this effect counteracts (stops) the linear ID increase.
Saturation region: Now the influence of the rising VDS dominates the width of the depletion region: Any further VDS increase causes the remaining n-channel to become smaller and smaller (pinch-off effect). As the result, both effects (ohmic behaviour and pinch-off) nearly cancel each other and ID remains nearly constant.
As we know, in electronics nothing is ideal - and this applies, of course, also to the FET: Up to a certain extend, the ohmic behaviour of the channel cannot completely stopped by the pinch-off effect. Hence, there is still a small ID increase with rising VDS values.