Solution:

We assume that there exists a matrix $B$ of the required kind with $B \neq A$. Now we consider in $A$ only those entries $a_{ij}$ that differ from the corresponding entries $b_{ij}$. There must be at least one such entry. All other entries of $A$ are struck out. Then, within each row or column, the number of remaining zeros equals the number of remaining ones, since these numbers are precisely interchanged in $B$ given equal row or column sums. Hence every number $x_{i}$ occurs in the remaining arrangement just as often as a summand of a sum $x_{i}+y_{j}<0$ as it does as a summand of a sum $x_{i}+y_{j} \geq 0$. The same holds for every number $y_{j}$.

Now we add up all the sums $x_{i}+y_{j}<0$ with $a_{ij} \neq b_{ij}$. The sum of these sums is necessarily $<0$. Likewise we add up all the sums $x_{i}+y_{j} \geq 0$ with $a_{ij} \neq b_{ij}$. The sum of these sums is necessarily $\geq 0$. But since each of the numbers $x_{i}$ and $y_{j}$ occurs equally often in both sums, these sums must have the same value — a contradiction! Therefore there can be no differing entries in $A$ and $B$, and hence $A=B$.