$f = \lfloor x \rfloor$, $f = \lceil x \rceil$, and constant functions.

We first verify that the above functions satisfy the problem's conditions. For all triples $(x, a, b) \in \mathbb{Q} \times \mathbb{Z} \times \mathbb{N}$, let
$$q = \lfloor \frac{x+a}{b} \rfloor.$$
It is easy to see that $bq \le x + a < b(q+1) \Rightarrow bq \le \lfloor x \rfloor + a < b(q+1)$, so
$$\lfloor \frac{\lfloor x \rfloor + a}{b} \rfloor = \lfloor \frac{x+a}{b} \rfloor.$$
Thus $f = \lfloor x \rfloor$ is indeed a solution. The other two solutions can be verified similarly.

Next we prove that these are all possible solutions. Consider the following two cases:

Case 1. There exists an integer $m$ such that $f(m) \neq m$.

Let $C = f(m)$. If $m > C$, then for all integers $y$, substituting $x = m, b = m-C, a = yb-C$ into the original equation gives $f(y) = f(y+1)$, so for all integers $y$, $f(y) = C$. If $m < C$, then instead taking $b = C-m$, we likewise obtain $f(y) = C$ for all integers $y$.

Now, for all rational numbers $y = p/q$ (where $q > 0$), substituting $(x, a, b) = (C - p, p - C, q)$ gives $f(y) = f(0) = C$. Hence $f$ is a constant function.

Case 2. For all integers $m$, $f(m) = m$.

Taking $b = 1$, we know that
$$f(x) + a = f(x + a) \quad (1)$$
holds for all $(x, a) \in \mathbb{Q} \times \mathbb{Z}$. Let $\omega = f(1/2)$.

Claim 1. $\omega \in \{0, 1\}$.

Proof. If $\omega \le 0$, substituting $(x, a, b) = (1/2, -\omega, 1-2\omega)$ gives $0 = f(0) = f(1/2) = \omega$. Similarly, if $\omega \ge 1$, take $(x, a, b) = (1/2, \omega-1, 2\omega-1)$.

Claim 2. For all rational numbers $0 < x < 1$, $f(x) = \omega$.

Proof. If not, there exists $p/q \in (0, 1)$ with the smallest denominator such that $f(p/q) \ne \omega$. It is easy to see that $\gcd(p, q) = 1$, and $q \ge 2$. If $q$ is even, then $p$ must be odd, so substituting $(x, a, b) = (\frac{1}{2}, \frac{p-1}{2}, \frac{q}{2})$ gives
$$f\left(\frac{\omega + (p-1)/2}{q/2}\right) = f(p/q) \ne \omega.$$
But by Claim 1, $\omega = 0$ or $1$, so the above equation in either case yields some $p'/q' \in (0, 1)$ satisfying $f(p'/q') \ne \omega$ with $q' < q$. This contradicts the assumption that $p/q$ has the smallest denominator, a contradiction.

Therefore $q$ must be odd; let $q = 2k + 1$. Substituting $(x, a, b) = (1/2, k, q)$ gives
$$f\left(\frac{\omega + k}{q}\right) = f\left(\frac{1}{2}\right) = \omega.$$
Since $p, q$ are coprime, there exist $r \in \{1, 2, \dots, q\}$ and an integer $m$ such that $rp - mq = k + \omega$; moreover, since the right-hand side is not a multiple of $q$, $r < q$. And if $m < 0$, then $rp - mq > q \ge k + \omega$, a contradiction, so $m \ge 0$. Similarly, if $m \ge r$, then $rp - mq < br - br = 0$. Hence, $0 \le m \le r - 1$. Now, taking $(x, a, b) = (\frac{k+\omega}{q}, m, r)$, we get
$$f\left(\frac{\omega + m}{r}\right) = f(p/q) \ne \omega.$$
This contradicts the minimality of $q$, a contradiction!

Now, if $\omega = 0$, then by Claim 2, $f(x) = \lfloor x \rfloor$. Similarly, if $\omega = 1$, then by Claim 2, $f(x) = \lceil x \rceil$. This completes the proof.