If there exists a pair of non-negative integers $\left(x_{0}, y_{0}\right)$ satisfying (1), then
$$a\left(x_{0}+1\right)+b\left(y_{0}+1\right)=a b,$$

then, we should have

and
thus
$$\begin{array}{l} a \mid b\left(y_{0}+1\right) \\ (a, b)=1 \\ a \mid y_{0}+1 \end{array}$$

Since $a$ and $y_{0}+1$ are both positive integers, we have $a \leqslant y_{0}+1$.
Similarly, we can prove: $b \mid x_{0}+1$, hence $b \leqslant x_{0}+1$. But in this case, the left side of (2) $\geqslant a b+b a=$ $2 a b>$ the right side of (2), which is a contradiction.

Therefore, (1) has no non-negative integer solutions.