No. For any $\epsilon$ and positive $x, P(x, \epsilon)>0$ and $P(x,-\epsilon) \leq 0$. Thus by continuity/IVT, $P(x, 0)=0$ for all positive $x$. Similarly $P(0, y)=0$ for all positive $y$. This implies $x y \mid P(x, y)$, and so we can write $P(x, y)=x y Q(x, y)$. But then this same logic holds for $Q$, and this cannot continue infinitely unless $P$ is identically 0 - in which case the conditions do not hold. So no such polynomial exists.