Obviously, $b \ne -2$ and $b \ne -1$. Adding $2$ to both sides of the equality, we get
$$\left(\frac{a+2}{b+1} + 1\right) + \left(\frac{a+1}{b+2} + 1\right) = 3 + \frac{6}{a+b+1},$$
hence
$$(a+b+3)\left(\frac{1}{b+1} + \frac{1}{b+2}\right) = \frac{3(a+b+3)}{a+b+1}.$$

Case 1. If $a + b + 3 = 0$, then every pair $(a, b) = (-3-u, u)$, where $u \in \mathbb{Z} \setminus \{-2, -1\}$, is a solution.

Case 2. If $a + b + 3 \neq 0$, then
$$\frac{1}{b+1} + \frac{1}{b+2} = \frac{3}{a+b+1},$$
which leads to
$$a = \frac{b^2 + 4b + 3}{2b + 3}.$$
Since $a \in \mathbb{Z}$, it results that $2b+3 \mid b^2+4b+3$. It follows that $2b+3 \mid 3$, so $b \in \{-3, -2, -1, 0\}$. We get one additional solution, namely $(a,b) = (1,0)$.