Proof
Let
$$I=\{n \mid 0 \leqslant n \leqslant a b-a-b, n \in \mathbf{Z}\}, A=a b-a-b,$$

and call a number representable if it can be written as $a x+b y(x, y \geqslant 0, x, y \in \mathbf{Z})$, otherwise it is called non-representable.

Clearly, $I$ contains
$$a b-a-b+1=(a-1)(b-1)$$

integers, so it suffices to prove that the number of representable numbers is equal to the number of non-representable numbers, i.e., to prove: for any $n \in I, n$ and $A-n$ are exactly one of which is representable.

First, $n$ and $A-n$ cannot both be representable. Otherwise, $A=n+(A-n)$ would also be representable, contradicting the conclusion of Example 3.

Second, if $n$ is non-representable, then $A-n$ is representable. Because if $n$ is non-representable, then the integers $x, y$ satisfying
$$n=a x+b y$$

must have exactly one of them negative. Without loss of generality, assume $x<0$. In this case, we can choose an appropriate $t \in \mathbf{Z}$ such that $y-a t \in[0, a), x+b t<0$ (otherwise $n$ would be representable). This means there exist $x<0, 0 \leqslant y<a$, such that $n=a x+b y$. Therefore,
$$\begin{aligned} A-n & =a b-a-b-a x-b y \\ & =a(-x-1)+b(a-1-y) \end{aligned}$$

Clearly, $-x-1 \geqslant 0, a-1-y \geqslant 0$, so $A-n$ is representable.
In summary, the proposition is established.