$f(x) = x+1$ Proof:

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There are three such $f$: (1) $f(x)=1$ for all $x$, (2) $f(x)=0$ if $x$ is odd and $f(x)=1$ if $x$ is even, and (3) $f(x)=x+1$ for all $x$. Let $P(x,y)$ denotes the given assertion. We obtain, by inspecting $P(0,0)$, that $f(0)=1$. Further, $P(x,1)$ gives $f(x+1)+f(x) = f(x)f(1)+1$. Here, taking $x=-1$ we get $f(-1)=f(-1)f(1)$, so $f(-1)=0$ or $f(1)=1$. First, suppose $f(1)=1$. Then, $f(x+1)+f(x)=f(x)f(1)+1$ gives $f(x+1)=1$ for all $x$, which is the solution in (1). Now assume $f(1)\ne 1$ and $f(-1)=0$. Taking $x=-2$ above, we get $f(-2)+f(-1)=f(-2)f(1)+1$, so $f(-2)(1-f(1))=1$. Moreover, $P(-1,-1)$ gives $f(-2)+f(1)=f(-1)^2+1$, so $f(-2)=1-f(1)$. Hence, $(1-f(1))^2=1$, yielding $f(1)\in 2$ or $f(1)=0$. If $f(1)=2$ then the recursion $f(x+1)+f(x)=f(x)f(1)+1$ gives $f(x+1)=f(x)+1$, yielding $f(x)=x+1$ for all $x$. Finally, if $f(1)=0$ then we easily get the second solution, which is easily verified to work for all $x,y$. The end.

Nice solution.