a lower bound for the maximum fraction of matchings not containing an edge
Your approach has been working well all along, only to miss the promising transformation at the last step.
Instead of
$$\frac{|E|-k}{|E|} \ge \frac{1}{|E|} \ge \frac{1}{n^2},$$
we can proceed as
$$\frac{|E|-k}{|E|} =1 - \frac{k}{|E|} \ge1- \frac k{k+1}=\frac1{k+1}.$$
Since $k\le n$, i.e., $\dfrac1{k+1}\ge\dfrac1{n+1}$,  we are not far from the goal $\dfrac1n$.
It takes a bit more effort to reach the goal. I will leave it to you to complete.



Here is a slightly easier proof. It shows, except possibly for graphs where all edges are separated, we can increase $1/n$ to $1/2$.
Consider whether there are incident edges in $G$.

There are incident edges. Say, edge $e_1$ and $e_2$ are incident. Every matching (of size $k$) can contain at most one of them. So either at least 1/2 of the matching of size $k$ does not contain $e_1$ or  at least 1/2 of the matching of size $k$ does not contain $e_2$. Note that $$1/2\ge 1/n.$$
There are no incident edges. Then there are at most $n$ edges. Since each matching does not contain at least one of the $|E|$ edges, we have , if $e$ is the edge that is absent from the most matchings of size $k$,$$\frac{m_{ne}}{m_k}\ge \frac{1}{|E|} \ge \frac{1}{n}.$$

$\checkmark$