By taking $b=1$, we get
$$f(f(a) + f(1)) = a,$$
for every integer $a$.
Let $n$ be an arbitrary positive integer. For $a = f(n) + f(1)$ and $b = 2f(1)$, we get $f(a) = n$ and $f(b) = 1$, and we have:
$$f(n+1) = f(n) + 3f(1) - 1.$$
Let $c = f(1)$. It is easy to prove by induction that for every integer $n$
$$f(n) = (3n - 2)c - n + 1.$$

Then $f(0) = -2c + 1$, and by taking $a = 0$ and $b = 1$ the initial expression becomes:
$$\begin{aligned} f(f(0) + f(1)) &= 0 \\ f(-c + 1) &= 0 \\ (3(-c + 1) - 2)c - (-c + 1) + 1 &= 0 \\ c(-3c + 2) &= 0. \end{aligned}$$
Because $c \in \mathbb{Z}$ we conclude $c = 0$. Therefore, the only solution of the given equation is $f(n) = -n + 1$. We can directly check that this function satisfies the given equation.