Hope i didnt do anything wrongly...
Put $y=0$, $f(f(x))=f(x)f(0)-f(x)+f(0)$
So $P(x,y): f(x-y)f(0)-f(x-y)+f(0)=f(x)f(y)-f(x)+f(y)-xy$
Let $f(1)=a, f(0)=b, f(-1)=c$.
$P(1,1): a^2=1+b^2$
$P(-1,-1): c^2=1+b^2$
$P(0,1): bc-c+b=ab-b+a$
Combining the first and second equations, we see that $a=c$ or $a=-c$.
If $a=c$ then from the third equation $-a+b=-b+a$ so $a=b$, which contradicts the first equation.
Else $a=-c$. From the last equation, $bc+b=0$.
If $b\ne 0$ then $c=-1$ and so $a=1$ and $f(1)=1$.
$P(2,1): 2f(0)-1=1-2$. So $f(0)=b=0$, contradiction.
Thus $b=0$ and $f(0)=0$.
$P(x,x): f(x)^2=x^2$.
If $f(c)=c$ for some $c$, then
$P(c,0)$ in the original equation (the one in the question):
$c=-c$, so $c=0$.
Thus the only solution is $f(x)=-x$ for all real $x$.