Will the voltage at A still be zero? Does it changes absolutely
nothing to have Vs2 there?
To find the answer, write an equation for the node voltage \$v_A\$ using superposition. For simplicity, assume the input resistance \$R_I\$ is infinite.
$$v_A = v_S \frac{R_2}{R_1 + R_2} + v_O \frac{R_1}{R_1 + R_2}$$
We also have
$$v_I = v_{S2}-v_A $$
$$v_O = Av_I $$
Thus, the first equation becomes
$$v_A = v_S \frac{R_2}{R_1 + R_2} + A(v_{S2}-v_A) \frac{R_1}{R_1 + R_2}$$
Rearranging yields
$$v_A = \frac{v_SR_2 + Av_{S2}R_1}{(1 + A)R_1 + R_2}$$
This is the general expression for \$v_A\$. As the gain \$A\$ gets very large,
$$v_A \approx \frac{Av_{S2}R_1}{AR_1} = v_{S2}$$
In the ideal case where the gain is 'infinite', this expression is exact. This is why we say there is a virtual short between the two input terminals of an ideal op-amp with negative feedback - the negative feedback ensures the inverting input voltage is identical to the non-inverting input voltage.
why can't electrons also flow from the negative terminal in battery 2, through the right bulb, and into the positive terminal of battery 1?
Because the only way to get to the negative terminal of battery 2 is to come from the body of the battery, and the only way to get there is from the positive terminal of the battery. And the only way to get there is through the switch in the friend's house, which is open so no current can flow through it.
Batteries don't create electrons from nothing. They "pump" them from one terminal to the other via a chemical reaction, causing current to flow (through the battery) from a low potential to a high potential.
Current only flows in complete circuit means the current has to flow through the battery just as much as it has to flow through every other circuit element.
As an aside, it's also why it's silly to say that current always flows from high potential to low --- the battery is a trivial example of when it's the other way around. And it's also silly to say that all current is a flow of electrons in the opposite direction from conventional current --- the battery is a trivial and everyday example of when ionic currents should be considered.
Best Answer
Both Martin and immibis have it correct. Here's a slightly more complete circuit, showing how I1 is absorbed by the op-amp output, then flows to the op-amps' negative supply voltage (the op-amp's positive supply voltage is not shown, but is still required).
You cannot accurately measure I1 at the -ve power supply, because there is additional quiescent current to bias the op-amp's internal transistors.