No, it won't pump two electrons to the negative terminus because there is a charge on the wire that is creating a field that opposes the motion of the electrons that wasn't there before. So something less than two electrons will be pushed onto the negative terminal. To keep things a bit more in perspective (while still using made-up numbers), before you connect the wire you might have 100,000 electrons on the negative terminal and a shortage of 100,000 electrons on the positive terminal. When you connect the wire you might get 100 electrons that flow onto the wire but only 90 electrons are moved from the interior of the battery to the negative terminus to take their place. At the same time that this is going on, 90 more electrons are removed from the positive terminal (remember, it is the positive plate of a capacitor!). You thus now have 100,090 electrons on the positive terminal, 99,990 electrons on the negative terminal, an 100 electrons on the wire.Ok. Now let's proceed.
Nothing is connected to the battery. The battery is at equilibrium. There are, say, 10 positive charges on its positive terminus, and 10 negative charges on its negative terminus. We connected a wire to the negative terminus. 2 electrons moved from the negative terminus into the wire. The battery pumped additional 2 electrons into the negative terminus.
The terminal voltage of the battery is not established by the number of electrons on the terminals -- a battery is NOT a capacitor!Now we have a new charge distribution between the battery's termini: there are 12 positive charges on the positive terminus and 10 negative charges on the negative terminus.
Do you believe that now the potential difference between the battery's termini is the same as it was at the equilibrium? Is it higher?