Solution 1. We first show that $f(f(x))$ must be affine. To see this, replace $(a, b)$ first by $(a - 1, a + 1)$ and then by $(a, a)$ to obtain
$$f(f(a - 1)) + f(f(a + 1)) = f(2a) = 2f(f(a)).$$
Therefore,
$$f(f(a + 1)) - f(f(a)) = f(f(a)) - f(f(a - 1)).$$
This means that there is a constant $m$ so that for all $a \in \mathbb{Z}$:
$$f(f(a + 1)) - f(f(a)) = m.$$
Inductively it follows that for all $a \in \mathbb{Z}$
$$f(f(a)) = ma + c.$$
Then, taking the original functional equation with $b = 0$.
$$f(a + 0) = f(f(a)) + f(f(0)) = ma + 2c.$$
Taking the full original function equation, then we have:
$$\begin{aligned} m(a + b) + 2c &= f(a + b) \\ &= f(f(a)) + f(f(b)) \\ &= f(ma + 2c) + f(mb + 2c) \\ &= m(ma + 2c) + 2c + m(mb + 2c) + 2c \\ &= m^2(a + b) + 2c(2m + 2). \end{aligned}$$
As this holds for arbitrary $a + b$ we must have:
$$\begin{aligned} m &= m^2 \\ 0 &= (2m + 1)c. \end{aligned}$$
The first equation implies either $m = 0$ or $m = 1$, so $2m + 1 \neq 0$ and the second equation then implies $c = 0$. Putting these together, we see that either $f(a) = 0$ for all $a \in \mathbb{Z}$ or $f(a) = a$ for all $a \in \mathbb{Z}$ and the possible values of $f(2020)$ are 0 or 2020.

Solution 2. Let $f^n$ denote the $n$-th iterate of $f$, i.e. $f^2(x) = f(f(x))$, $f^3(x) = f(f(f(x)))$ etc. The given equation can then be written as
$$f(a + b) = f^2(a) + f^2(b).$$
Using $a = f(0)$ and $b = 0$, we get $f^2(0) = f^3(0) + f^2(0)$, which implies
$$f^3(0) = 0.$$
Let $a = b = 0$ to get $f(0) = 2f^2(0)$. Let now $a = b = f^2(0)$ and use the two previous identities to see that
$$f^2(0) = f(f(0)) = f(2f^2(0)) = 2f^4(0) = 2f(f^3(0)) = 2f(0) = 4f^2(0).$$
This implies $f^2(0) = 0$. Letting $b = 0$ in the original equation, we now obtain $f^2(a) = f(a)$ for all $a \in \mathbb{Z}$. This means that the given equation is equivalent to Cauchy's equation
$$f(a + b) = f(a) + f(b)$$
whose solutions over the integers are known to be of the form $f(a) = ca$ for some integer $c$. From $f^2(1) = f(1)$ we obtain $c^2 = c$, i.e. $c = 0$ or $c = 1$. Therefore, the possible values of $f(2020)$ are 0 or 2020.