Counterexample shown here for just adding edges:
If $k=5$, then you'd need another $3$ edges from each of the vertices in $\{5,6\}$. Because $5$ and $6$ already have edges to $11$ and $12$, all of those additional edges must be to vertices in $\{7,8,9,10\}$. But all vertices in $\{7,8,9,10\}$ already have degree $4$, and thus can only accept one edge each. That means that only a total of $4$ edges from $\{5,6\}$ can be accepted, leaving $2$ stubs on $\{5,6\}$ with no place to go.
It might still be possible to extend the graph if you add both edges and vertices, but I'm not sure how to systematically do that.
Incorrect old answer:
My original answer was incorrect as highlighted by Aryabhata's comment (unless you allow for weighted graphs or multigraphs), but I've left it below for reference.
Succinctly, yes. You've basically already answered the question.
I'll try to flesh it out a bit to make it more evident, but really I'm just repeating your own intuition:
Given a bipartite graph $G=(X,Y,E)$, $$ \sum_{x \in X} d(x) = \sum_{y \in Y} d(y) = |E| $$
since each edge counts towards the degrees of one vertex in both $X$ and $Y$.
We have assumed $G$ is balanced, so let $m \equiv |X| = |Y|$. Furthermore, we assume $k < m$.
Since $d(i) \le k, \forall i \in X \cup Y$, $|E| \le km$.
Case 1: $|E| = km$. Then $d(i) = k$, so done.
Case 2: $|E| < km$. Then for every vertex $i$, put $k - d(i)$ stubs on $i$ (recall that a stub is half of an edge, two stubs on different vertices can be joined to form an edge between those vertices, and that a stub counts towards the degree of a vertex).
Now, all vertices have degree $k$. However, we still need to join the stubs together to form edges to turn this back into a bipartite graph. Note that we added a total of $km - |E|$ stubs to vertices in $X$. Order them $\{s_1, ..., s_{km-|E|}\}$. Similarly, we can order the stubs in $Y$ as $\{t_1, ..., t_{km-|E|}\}$.
Then just join together $s_j$ and $t_j$ for $j \in [1,km-|E|]$, and you have your new regular balanced bipartite graph $G'$.
Best Answer
By the definition of degeneracy, it is the minimum number $k$ such that the graph is $k$-degenerate. Certainly, $K_{m, n}$ is $max \{m, n\}$-degenerate. But in fact we have $\delta(K_{m, n}) = min \{m, n\}$. WLOG, let us assume that $m \geq n$ then we need to show that $G$ has degeneracy $n$. Let's say the left part consists of $m$ nodes and the right part consists of $n$ nodes, then if the subgraph contains any left node, then each left node has degree at most $n$, and if the subgraph contains no left node, then it contains no edges at all. And clearly, the minimum node degree in the original graph is $n$ so the degeneracy cannot be strictly less then $n$, completing the proof.