what is there to stop the electrons from the negatively charged
capacitor plate 'flowing back' to where the negative terminal of the
battery was located before?
The electric field that exists between the plates of the capacitor due to the separation of charge. The electrons on the negatively charged plate are attracted to the positively charged plate. It's actually a minimum energy configuration.
Also, while the battery is connected to the charged capacitor, the negative plate, connecting wire, and negative terminal of the battery are all negatively charged. Similarly for the positive plate, connecting wire, and positive terminal. Indeed, there is an electric field between the connecting wires that gives rise to the voltage across them.
When the battery is disconnected, an ideal voltmeter connected across the connecting wires will continue to read the battery voltage since that is the voltage the capacitor and connecting wires charged to. Electrons didn't need to leave the negative plate to establish this voltage - the charge distribution giving rise to this voltage was already there.
With the edit to the question which added diagrams, I can now more easily make a point that I only touched on in my initial answer. Note: I'll be leaving out what I feel to be unnecessary complications in order to make the essential point.
The two leads that connect to the capacitor plates are conductors and, as such, form a capacitor in parallel with the primary capacitor formed by the plates. This stray or parasitic capacitance is almost always negligible compared to the primary capacitance but, particularly in RF circuits, it must sometimes be taken into account. This 'inter-lead' capacitance is of course dependent on the geometry, e.g., lead spacing, length etc.
Regardless, this capacitor is also charged by the battery to the battery voltage and this means that the wires connected to the plates have a small amount of charge density to establish the electric field between the leads and thus the voltage across.
When the battery is disconnected (in some 'ideal' way that doesn't disturb the leads etc.), this stray capacitance remains charged (again, ignoring all the physical ways charge might 'leak' from one lead to the other).
That is to say, it is not my understanding that electrons from the negative plate charge this stray capacitance after the battery is disconnected, it was already charged.