very easy. Putting $y=1$ we get $f(x+1)=f(x)+1$ So by iteration $f(x+n)=f(x)+n\forall n\in\mathbb Z$ and also $f(x)=x+1\ \forall x\in\mathbb Z$. Now for $x=\frac pq$ with $p,q\in\mathbb Z$, we have $p+1=f(p)=f(qx)=(q+1)f(x)-f(x+q)+1=qf(x)-q+1$, thus $f(x)=\frac {p+q}q=x+1\ \forall x\in\mathbb Q$. That's all.

spanferkel wrote: $f(qx)=qf(x)-q+1$ How did you get that? $f(qx)=(q+1)f(x)-f(x+q)+1,$ but you can't imply $f(x+q)=f(x)+q,$ because you didn't proof that for $x\in\mathbb{Q}.$

I was just editing that. I think now it should be clear. your last line is already proved for $q\in\mathbb Z$.

Now excellent

Solution copied over here. $P(x,0): f(0)=f(x)f(0)-f(x)+1$. $P(x,1): f(x)=2f(x)-f(x+1)+1\implies f(x+1)=f(x)+1$. So $f(x)=x+1\forall x\in \mathbb{Z}$. For rational $\frac{p}{q}$, where $p$ and $q$ are integers, $P\left(\frac{p}{q},q\right): f(p)=f\left(\frac{p}{q}\right)f(q)-f\left(\frac{p}{q}+q\right)+1$. Thus, $p+1=f\left(\frac{p}{q}\right)(q+1)-f\left(\frac{p}{q}\right)-q+1$. So we have $qf\left(\frac{p}{q}\right)=p+q\implies f\left(\frac{p}{q}\right)=\frac{p}{q}+1$. Thus, $\boxed{f(x)=x+1}$, which works.