The two solutions are $f(x) = x$ and $f(x) = -x$. We prove this in three stages. First we show that $f$ is self-inverse, that is, $f(f(x)) = x$ for all integers $x$. Secondly we show that $f$ is additive. Thirdly, we demonstrate that the stated solutions are the only self-inverse additive functions.

To show the self-inverse property, interchange $a$ and $b$ in the original equation
$$f(b + f(a)) = a + f(b),$$
then apply $f$ again to each side, which gives
$$f(f(b + f(a))) = f(a + f(b)) = b + f(a).$$
Any integer can be represented as $b + f(a)$ hence $f(f(x)) = x$ for all $x$. For additivity, let $c = f(b)$. By the self-inverse property, any integer $c$ can be written in this form, setting $b = f(c)$. The original equation becomes
$$f(a + c) = f(c) + f(a).$$
Putting $a = c = 0$ implies $f(0) = 0$. It follows by induction that $f(x) = x f(1)$ for positive integers $x$. Writing $c = -a$ proves that $f(x) = x f(1)$ for all negative $x$. Therefore, $f(x)$ is linear with slope $f(1)$ and $y$-intercept of zero. Finally, the self-inverse property with $x = 1$ gives $1 = f(f(1)) = f(1)^2$, whence $f(1)^2 = 1$ and $f(1) = \pm 1$ yielding the two solutions claimed. It is straightforward to check that both indeed are solutions.