A closed surface S is set up around a conducting wire that is...

According to Gauss's law, the electric flux through a closed surface is given by $\Phi = \dfrac{q_{\text{enclosed}}}{\epsilon_0}$.

Initially, when the switch is open, the wire is electrically neutral, so the net charge enclosed by the surface $S$ is zero ($q_{\text{enclosed}} = 0$). Thus, the initial electric flux is zero.

When the switch is closed, current starts flowing. The problem explicitly states that the number of electrons entering the closed surface $S$ equals the number leaving it during any given time interval. This means there is no net accumulation or depletion of charge inside the surface $S$.

Therefore, the net charge enclosed by the surface $S$ remains zero ($q_{\text{enclosed}} = 0$) even after the switch is closed.

Consequently, the electric flux through the closed surface $S$ remains zero. Since the flux was zero initially and continues to be zero, it remains unchanged.