Would separating $A$ into two nodes connected by one edge work? Also, the minimum degree of your graph is $2$ at each of the four corner nodes.
The graph with the largest minimum degree, lowest edge connectivity, and lowest vertex connectivity is formed by connecting two complete graphs with one singular edge like a bridge.
Let $n$ be any positive integer. Induction over $m$.
$m=1$: $K_{1,n}$ is the star, which is connected, so $\lambda(K_{1,n})\geq 1$. But we can remove any single edge to get an isolated vertex, so $\lambda(K_{1,n}) = 1$.
$m\mapsto m+1$: Regard $H\cong K_{m,n}$ as a specific subgraph of $K_{m+1,n}$. Denote the vertices in the partition of $K_{m+1,n}$ by $v_1,\ldots,v_{m+1}$ and $u_1,\ldots,u_n$ respectively. Let $E$ be an edge-cut of $K_{m+1,n}$, then $E\cap E(H)$ is an edge-cut of $H$.
First case: $n\leq m$. By induction thesis $|E\cap E(H)|\geq n$ and so $|E|\geq n$. Since we can specify an edge-cut of size $n$ we have $\lambda(K_{m+1,n})=n$.
Second case: $n>m$. Parallel to the first case, we have $|E\cap E(H)|\geq m$ and so $|E|\geq m$. If one of the edges incident with the new vertex $v_{m+1}$ was in $E$ (but because of the incidence not in $E(H)$), we would have $|E|\geq m+1$. On the other hand assume that none of the edges incident with $v_{m+1}$ is in $E$, so all of $u_1,\ldots,u_n$ are adjacent to $v_{m+1}$. Since $E$ is edge-cut, there must be a vertex in $\{v_1,\ldots,v_m\}$ that isn't adjacent to any of the $u_1,\ldots,u_n$ vertices (else $G\setminus E$ would still be connected). Therefore, at least all $n$ edges in that vertex must be cut and we have $|E|\geq n\geq m+1$, which concludes the proof, since we can specify an edge-cut of size $m+1$.
Best Answer
Without loss of generality, assume that $m \le n$. And we'd like to show that the vertex-connectivity $\kappa(G) = m-1$, where $G$ is the complete bipartite graph on $m + n$ vertices with one edged, say $(a,b)$, removed.
It is enough to show that removing $m-2$ vertices cannot make $G$ disconnected. Suppose the left and the right partite set are of size $m, n$ respectively. After removing $m-2$ vertices from $G$, we get a complete bipartite graph with at most one edge removed. Suppose the partite sets are $A$ and $B$ of the resulting graph. We also know that $|A|, |B| \ge 2$.
Take any two vertices $u$ from $A$ and $v$ from $B$. As $|A|, |B| \ge 2$, we can find $u'\in A\setminus \{a\}$, and $v'\in B\setminus \{b\}$. Thus $u\sim v'\sim u' \sim v$, and so $A$ and $B$ are connected.