We are given $a$ and $b$, find all functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$ such that:

$$
f(x+a+f(y))=f(x+b)+y
$$
We fix $x$, and we have the bijectivity of $f$.

We take $x=-b$, which gives $f(a-b+f(y))=f(0)+y$. On the other hand, we set $y=f(0)$, we have:

$$
f(x+a+f(f(0)))=f(x+b)+f(0)=f(a-b+f(f(x+b)))
$$

Then by injectivity,

$$
f(f(x))=f(f(0))+x
$$

We can now set $x=f(w)-b$ and we get

$$
f(f(w)+f(y)+a-b)=w+y+f(f(0))=f(f(w+y))
$$

By injectivity, we obtain:

$$
f(w)+f(y)+a-b=f(w-y)
$$

By setting $g=f+(a-b)$, we get the Cauchy equation on $g$ in $\mathbb{Q}$. Therefore, $g$ is linear, and by synthesis, we get: $f(x)=x+b-a$ or $f(x)=-x+b-a$