We rewrite the system as
$$F(x) > 0 \land G(x) < 0,$$
where $F(x) = x^2-(a-1)x-b$ and $G(x) = x^2-ax-b+1$. Observe that $F(x)-G(x) = x-1$.
The condition $a+b>2$ implies that
$$F(1) = G(1) = 2 - a - b < 0,$$
hence $x=1$ is not a solution. However, $G(1) < 0$ implies that the quadratic equation $G(x) = 0$ has a root $x_0 > 1$. Then
$$F(x_0) = F(x_0) - 0 = F(x_0) - G(x_0) = x_0 - 1 > 0.$$
From $F(1) < 0$ and $F(x_0) > 0$ we deduce that there exists a root $x_1$ of $F(x) = 0$ that belongs to the open interval $(1, x_0)$. Since
$$F(1) < 0 \land F(x_1) = 0, \quad \text{and} \quad G(1) < 0 \land G(x_0) = 0,$$
any $x \in (x_1, x_0)$ is a solution to the original system.