$\left(P(x)e^{-x^2}\right)'=0$ must be satisfied by $0$, $\pm a$, $\pm b$, and no other values of $x$.
I have a solution that I can't find a problem with, but I get the wrong answer:
$\left(P(x)e^{-x^2}\right)'=e^{-x^2}\left(P'(x)-2xP(x)\right)=0\quad\iff\quad g(x):=2xP(x)-P'(x)=0$. A couple of things we can note immediately: $g(x)$ is a polynomial with $5$ distinct real solutions, and therefore has degree $\geq 5$. Also, $g$ is even in $x$, and therefore has only even exponents (i.e. is a polynomial in $x^2$), and so has degree $\geq 6$. Also, it has no constant term, since $g(0)=0$. $P(x)$ must have degree $\geq 5$, and, when choosing $P(x)$, we may assume that it is monic, since scaling $P$ by a constant factor scales its derivative, and therefore $g(x)$, by the same factor, which has no effect on the zeros of $g$. Let $$P(x)=x^5+Ax^4+Bx^3+Cx^2+Dx+E;$$ $$P'(x)=5x^4+4Ax^3+3Bx^2+2Cx+D;$$ $$g(x)=2x^6+2Ax^5+(2B-5)x^4+(2C-4A)x^3+(2D-3B)x^2+(2E-2C)x-D.$$ Employing the assumptions that $g$ is even and has no constant term, $A=C=D=E=0$, so $P(x)=x^5+Bx^3$, and $g(x)=2x^6+(2B-5)x^4-3Bx^2$. Suppose that $g$ has the required zeros, then $g(x)=2x^2(x^2-a^2)(x^2-b^2)$, so $B=-2a^2b^2/3$ (comparing the coefficients of $x^2$). Hence, $P(x)=x^5-2a^2b^2x^3/3$. This is not correct. We can see that something is wrong, since comparing the coefficients of $x^4$ yields $5-2B=5+4a^2b^2/3=2(a^2+b^2)$, which doesn't hold for all $a$ and $b$.