You can bound $\lvert N(S)\rvert$ from below by considering the worst case that all neighbours have the maximal degree, $n-d+1$. Then each vertex in $S$ of degree $n-d+1$ has as many outgoing edges as a worst-case neighbour has incoming edges, so the "cost balance" of such a vertex is neutral. Each vertex in $S$ of degree $n-d$ "saves" one edge, but there can be at most $n-d$ of these in $S$, so only $n-d$ edges can be "saved", whereas $n-d+1$ edges would have to be "saved" to "save" one worst-case neighbour. So no neighbour can be "saved" even in the worst case, and thus there must be as many neighbours as vertices in $S$.
I can write that out in formulas if you want, but you said you wanted a hint.
[Edit in response to request for formulas:]
Since all edges incident at a vertex in $S$ are also incident at a vertex of $N(S)$, we have
$$\sum_{w\in N(S)}\deg(w)\ge\sum_{v\in S}\deg(v)\;.$$
If $S$ contains $n_-$ vertices of degree $n-d$ and $n_+$ vertices of degree $n-d+1$, then
$$\begin{eqnarray}
\sum_{v\in S}\deg(v)&=&n_-(n-d)+n_+(n-d+1)\\\
&=&(n_-+n_+)(n-d+1)-n_-\\\
&=&\lvert S \rvert (n-d+1)-n_-\\\
&\ge&\lvert S \rvert (n-d+1)-(n-d)\;,
\end{eqnarray}$$
since there are only $n-d$ vertices of degree $n-d$. On the other hand, since no vertex has degree more than $n-d+1$, we also have
$$\sum_{w\in N(S)}\deg(w)\le \lvert N(S) \rvert (n-d+1)\;.$$
Putting this all together gives
$$\lvert N(S) \rvert (n-d+1) \ge \lvert S \rvert (n-d+1)-(n-d)\;,$$
$$\lvert N(S) \rvert \ge \lvert S \rvert -\frac{n-d}{n-d+1}\;.$$
But $\lvert N(S) \rvert$ and $\lvert S \rvert$ are integers and the last term is less than $1$, so we can drop the last term to conclude that $\lvert N(S) \rvert\ge\lvert S \rvert$. Linking this back to the original answer, each vertex in $S$ of degree $n-d$ can change the balance in the penultimate inequality by at most $1$, and there are at most $n-d$ such vertices, so the balance can only be changed by $n-d$, but it would have to change by $n-d+1$ to make a difference in the last inequality, where the division by $n-d+1$ corresponds to the fact that we need to "save" $n-d+1$ edges in order to "save" one worst-case neighbour.
I think you have the correct idea. Maybe we can formalize it little more.
Let $G$ be a bipartite graph and $X$ and $Y$ are bipartitions of $V(G)$. Also let $x\in X$ be a vertex with degree $1$ and $e = xy$ be the edge incident to $x$. So, the edge that is connected to $x$ is connected to $y \in Y$. Then the claim should be there exists a maximal matching containing $e$ because we can find an example where there is a maximal matching does not contain $e$ with the same number of edges (but your claim is just to include $e$ to maximal matching safely so I don't think this is important here).
Now, let $M$ be the maximal matching with $|M|$ edges including $e$. Suppose for a contradiction that there exists a matching $M'$ with $|M'| \ge |M|+1$. Since $M$ is the maximal matching including $e$, $M'$ cannot include $e$. Therefore, only possibility for this matching is to include $ay$ as an edge where $a \in X$. Then we can remove $ay$ and add $e=xy$ to $M'$ without changing $|M'|$. But then our new matching is a matching that includes $e$ and with size $|M'| > |M|$, contradictory to the assumption that $M$ is the maximal matching including $e$.
Best Answer
Add more edges (and maybe more vertices) until the bipartite graph is $\Delta(G)$-regular. What can you say about regular bipartite graphs?