There are two cases.  In your solution the two cases are not clearly separated, probably because you didn't realize the two cases.  Let me explain the two cases, hopefully more clearly.

Suppose we already have the number of mismatches for number of pairs below $n$ (that's including $1$ pair through $n-1$ pairs), and that's $a_1, a_2, \ldots, a_{n-1}$.  Now let's consider a new pair, $W_n$ and $M_n$.  The requirement is that they are not paired up.  So let's pair up $W_n$ and $M_j$ for some $j$ (that's $n-1$ cases, but we will multiply that later).  Now, who is pairing up with $W_j$ then?  Here we get two cases.

Case 1: $W_j$ is paired with $M_n$.  Here we have the $n$th pair swapped with the $j$th pair.  The remaining $n-2$ pairs will have $a_{n-2}$ ways of mismatching.

Case 2: $W_j$ is not paired with $M_n$.  Here, we pretend that $M_n$ takes the place of $M_j$ and we still have the pairs $1$ through $n-1$, and that's $a_{n-1}$ ways of mismatching.

Now we add up the two cases, and multiply each case by $n-1$ to get

$$a_n = (n-1)a_{n-1} + (n-1)a_{n-2}.$$