The result holds for constant and linear polynomials. Suppose $d \ge 2$ is the degree of $P$. WLOG, let $P$ be monic. Write $P(x)=x^d+\sum^{d-1}_{i=0} c_ix^i$. Then we know that there exists sequence $(x_n)_{n \ge 1}$ and $(y_n)_{n \ge 1}$ of integers such that $P(y_n)=-P(x_n)$. WLOG, let $x_n>0$ for all $n>M$ for some positive constant $M$. It is clear that $d$ is odd otherwise $P(x)>0$ for all $|x|$ large enough. For two functions $f,g$ we say that $f \asymp g$ if $$\lim_{ n \rightarrow \infty} \frac{f(n)}{g(n)}=1.$$Note that $P(x) \asymp x^d$. From $d$ odd and $$x_n^d \asymp P(x_n)=-P(y_n) \asymp -y_n^d,$$we have $x_n \asymp -y_n$. Notice that as $d$ is odd and $P(x_n)+P(y_n)=0$ we get $$x_n+y_n=\frac{x_n^d+y_n^d}{\sum_{i+j=d-1} x_n^i(-y_n)^j}=\frac{-\sum^{d-1}_{i=0} c_i(x_n^i+y_n^i)}{\sum_{i+j=d-1} x_n^i(-y_n)^j}, $$which converges as $x_n \asymp (-y_n)$. As a convergent sequence of integers is eventually constant, we see that for all $n$ large enough, there exists $c$ such that $x_n+y_n=c$. Thus, the equation $P(c-x)=-P(x)$ holds true for infinitely many values of $x$, so it holds for all $x$. Point $\left(\frac{c}{2},0\right)$ is the center of symmetry for the graph of $y=P(x)$ as desired. $\, \square$