Let $g(x)=f(x)-1$. Plugging in $f(x)=g(x)+1$ to $P(x,y)$, we obtain \[(x-y)g(x+y)-(x+y)g(x-y)=2y(g(x)-g(y)).\]By abuse of notation, we will refer to this as $P(x,y)$ henceforth and solve for $g$. $P(x,x)$ yields $-2xg(0)=0$ so $g(0)=0$. Then $P(x,-x)$ yields $0=-2x(g(x)-g(-x))$ so $g(x)=g(-x)$ for all $x\neq 0$. This is clearly true for $x=0$ as well, so $g$ is even. Now $P(x,y)$ and $P(y,x)$ and evenness yield \[x\cdot[(x-y)g(x+y)-(x+y)g(x-y)]=2xy(g(x)-g(y))=-2yx(g(y)-g(x))=-y\cdot[(y-x)g(x+y)-(x+y)g(y-x)]=-y\cdot[(y-x)g(x+y)-(x+y)g(x-y)],\]so \[(x-y)^2g(x+y)=(x+y)^2g(x-y).\]Changing variables yields $y^2g(x)=x^2g(y)$ for all $x,y\in\mathbb{R}$, i.e. $g(x)/x^2=g(y)/y^2$ for all $x,y\neq 0$. So $g(x)=cx^2$ for all $x\neq 0$ for some constant $c$. Furthermore, this holds for $x=0$ as well by $g(0)=0$. So, we obtain solutions of the form $f(x)=cx^2+1\forall x\in\mathbb{R}$ for each $c\in\mathbb{R}$, and we easily see that these work.