When we talk about exchanging electron $i$ with electron $j$, we are actually changing the wavefunction according to
$$\Psi(..., x_i, ..., x_j, ...) \to \Psi(..., x_j, ..., x_i, ...).$$
The operation is taken by the parity operator $P$. Applying it twice would
return the wavefunction to its original form. So the following eigenvalue equation is satisfied
In short: The presence of the $\sigma_h$ plane you see depends on the conformation of the methyl groups.
If your example were either benzene or 1,4-dichlorobenzene, then you were be right, the molecule would contain a plane of symmetry identical to the plane defined by the benzene ring. This $\sigma_h$ plane then could be described as in the following ...