Electronic – 5V digital output to 24V

If you look at the characteristic frequency of the hump it's on the order of 100's of KHz. the only thing n that circuit that has a dominant pole in that range will be the power supplies. Look at the lower rail and see if it correlates to the hump.

You are misinterpreting what the datasheet says. Fortunately.

Vout low is the voltage at specified current when load is from output pin to Vdd and the port is pulling the pin low against the pullup load.

When you are driving a ground referenced divider the load is from pin to ground. The port is "perfectly happy" to turn off essentially completely and allow the load to pull the pin to ground - but it will in fact (probably) help with an active pulldown. .

There MAY be some issues with bias and leakage currents (usually associated with input modes) if you are aiming for the last few microvolts, but in your application it will reach true-enough ground.

Datasheet (which you should always provide a link to) is at http://ww1.microchip.com/downloads/en/DeviceDoc/40001723C.pdf

See ~= page 251

RELATED:

Worst case specifications should be used for design BUT the conditions under which they are specified should be noted and where these do not match your conditions suitable adjustments may be made. With due care :-).

If you did care about Voutlow then you also need to note the conditions under which it is specified.

P251 in the datasheet says Vout low max = 0.6V (as you note). BUT they also state the loading conditions under which this is true.

AT 8 mA with Vdd = 5V = (5V-0.6V) / 8 mA = 550 Ohm load or
6 mA at Vdd = 3.3V = (3.3-0.6)/100 Ohm load or 450 Ohm load or
1.8 mA at Vdd = 1.8V = (1.8-0.6)/1.8 mA or 667 Ohm load.

While port pin resistance is NOT linear with load it is liable to be somewhat linear with load.

If you use higher load resistors (to Vdd) than above you can expect approximately proportionately lower Vout.