Now I see you have modified your question, and are referring to discharge time. I've edited my answer with an update further down.
I believe that you are seeing the "rise time" of your power supply, when you switch it on. There's no way that supply has an output impedance in the Megohms. Typically you'd expect an output impedance near zero.
If you switch the supply on prior to applying its 5V directly across the capacitor (using a switch, for example), letting its output settle to 5V first, then you'd see capacitor voltage rise almost instantaneously, not over many, many milliseconds.
The supply's own power switch is applying mains voltage to its own internal circuitry, and the time between that switch's closure and the full 5V appearing at the supply's output terminals is going to have more to do with the supply's own internal architecture than whatever load it is powering.
For instance, a power supply that employs a full bridge rectifier and beefy reservoir capacitor might take several mains cycles to fully charge that capacitor, for a delay in the tens of milliseconds.
If the supply contains a slew rate limited output, or current limiting, perhaps what you are witnessing is this in action. Maybe there's a microcontroller in there that takes 90ms to boot.
There are a dozen possible reasons for the rise time to be so long, but The "R" in an RC circuit (where C=100nF) is not one of them.
For a 100nF capacitor to discharge by 2.5V in 90ms requires a current of about:
$$ I = C\frac{dV}{dt} \approx 100nF \times \frac{2.5V}{90ms} \approx 3\mu A $$
I presume this is a ceramic capacitor, whose self-discharge current will be picoamps, or at worst nanoamps, so that's unlikely to be the cause of such a rapid discharge. I would expect a good quality 100nF ceramic capacitor to self-discharge over many seconds, without any load across it.
Dielectric absorption is going to have some effect, but not this significant, so somehow 3μA is passing.
It could be a really poor quality capacitor, or a badly labelled one.
It could be you, holding the probe against the capacitor leads with your fingers. It could be finger grease and sweat bridging the capacitor's leads after you've man-handled the thing. It could be any number of contaminants on the breadboard or structure you are using to build the experiment.
While the above reasons are likely to have an effect, the biggest contributor is likely to be the probe, which as you pointed out should have a resistance of 10MΩ. To isolate the probe, you'll need a FET input voltage follower:
simulate this circuit – Schematic created using CircuitLab
The op-amp will draw negligible current from the capacitor, while faithfully reproducing capacitor voltage at its output, so (assuming the circuit is free from contaminants) any voltage droop shown on the oscilloscope will chiefly be due to capacitor self-discharge.
