$P(x,0)\implies f(x)=1,\quad\forall x\in\mathbb{R}$, which indeed fits.

$P(2,2)\implies f(2)^{2}=1$ f(2)=-1$P(1,1)\implies f(1)^{2}-f(1)-2=0$ f(1)=2$P(x,1)\implies f(x+1)+f(x)=1,\quad\forall x\in\mathbb{R}\implies f(x+2)=f(x),\quad\forall x\in\mathbb{R}$ $P(x,y+2)-P(x,y)\implies f(xy+2x)=f(xy),\quad\forall x,y\in\mathbb{R}$ $x:=\frac{1}{2},y:=2\implies f(2)=f(1)$, contradiction. f(1)=-1$P(x,1)\implies f(x+1)=2f(x)+1,\quad\forall x\in\mathbb{R}$ $P(x,y+1)-2P(x,y)\implies f(x)=f(xy+x)-2f(xy)-2,\quad\forall x,y\in\mathbb{R}$ $y:=-1\implies f(x)+2f(-x)=-3,\quad\forall x\in\mathbb{R}$ $x:=-x\implies f(-x)+2f(x)=-3,\quad\forall x\in\mathbb{R}$ $\implies f(x)=-1,\quad\forall x\in\mathbb{R}$, which indeed fits. f(2)=1$P(1,1)\implies f(1)^{2}-f(1)=0$ f(1)=1$P(-1,1)\implies f(0)=1$, contradiction. f(1)=0$P(x,1)\implies f(x+1)=f(x)+1,\quad\forall x\in\mathbb{R}$ $P(x,y+1)-P(x,y)\implies f(x)=f(xy+x)-f(xy)-1,\quad\forall x,y\in\mathbb{R}$ $y:=\frac{y}{x}\implies f(x)+f(y)=f(x+y)-1,\quad\forall x\neq 0,y\in\mathbb{R}$. Since $f(0)=-1$, $f(x)+f(y)=f(x+y)-1,\quad\forall x,y\in\mathbb{R}$ Let $g(x)=f(x)+1,\quad\forall x\in\mathbb{R}$. Then $g(x+y)=g(x)+g(y),\quad\forall x,y\in\mathbb{R}$. From $f(x+y)+f(x)f(y)=f(xy)+1$ we have $f(x)+f(y)+f(x)f(y)=f(xy)$, implies $g(xy)=g(x)g(y),\quad\forall x,y\in\mathbb{R}$. From here, it's easy to prove that $g(x)=x,\quad\forall x\in\mathbb{R}$ and $f(x)=x-1,\quad\forall x\in\mathbb{R}$, which indeed fits.

All solution: $f(x)=-1,\quad\forall x\in\mathbb{R}$, $f(x)=1,\quad\forall x\in\mathbb{R}$, $f(x)=x-1,\quad\forall x\in\mathbb{R}$.