If $\deg P(x) = 0$ then $P(x) = 0$, and the statement of the problem is obviously true. Consider the case $\deg P(x) > 0$. Without loss of generality, we can assume that the leading coefficient of $P(x)$ equals $1$. If the degree of $P(x)$ is even then for $x$ with absolute value big enough, we will have $P(x) > 0$ and as a consequence, we can find only finitely many pairs of integers $m, n$ for which $P(m) + P(n) = 0$, which contradicts the given condition. So, the degree of $P(x)$ should be odd (we will use this fact further).

Now, we note that from some $x$ the polynomial $P(x)$ strictly increases and tends to infinity when $x$ tends to infinity (if $P'(x) > 0$ then this property holds for all $x$, if the equation $P'(x) = 0$ has real roots then we could choose $x$ greater than the biggest root of $P'(x)$). Moreover, for each integer $n$, there exists finitely many integers $m$ for which $P(m) = -P(n)$ (a polynomial can admit one value only in a finite number of points, not exceeding its degree).

With the above argument, we see that for all $C$, there is a pair of integers $m, n$ such that $P(m) + P(n) = 0$, where $m$ and $n$ have opposite signs and have absolute values greater than $C$.

Assume that $P(x)$ has degree $k$ and $P(x) = x^k + a x^{k-1} + \dots$ (here $\dots$ denotes the lower terms). It is easy to choose a number $d$ such that the polynomial $P(x-d)$ has the form $x^k + b x^{k-2} + \dots$, i.e., the coefficient of $x^{k-1}$ equals $0$.

Indeed, $P(x-d) = (x-d)^k + a(x-d)^{k-1} + \dots = x^k - k d x^{k-1} + a x^{k-1} + \dots$ so, we just need to choose $d = \frac{a}{k}$. Now, we prove that the point $(d; 0)$ is the center of symmetry of the graph of $y = P(x)$.

Put $P(x-d) = Q(x)$, we will prove that $Q(x) = -Q(-x)$ for all real $x$.

As we know, $Q(x) = x^k + b x^{k-2} + \dots$ and the equation $Q(m) + Q(n) = 0$ has infinitely many solutions for which $m-d, n-d$ are integers. We choose the solutions with big enough absolute value with $m > 0, n < 0$. We will prove that $|m| = |n|$.

Indeed, assume that $|m| < |n|$. Consider the case $n = -m - 1$ (other cases can be proved similarly). Then
$$Q(m) + Q(n) = Q(m) + Q(-m-1) = \\ m^k + b m^{k-1} + \dots + (-m-1)^k + b(-m-1)^{k-2} + \dots = -k m^{k-1} + R(m)$$
where $R(x)$ is a fixed polynomial of degree not greater than $k-2$. If $m$ is big enough, the value of $k m^{k-1}$ will be bigger than $|R(m)|$ so the sum $-k m^{k-1} + R(m)$ will be less than $0$. When the absolute value of $n$ increases then the sum $Q(m) + Q(n)$ decreases (because $Q(n)$ is decreasing). So, it is not possible to have $|m| < |n|$ where $m, n$ have big enough absolute value. Similarly, we cannot have the case $|m| > |n|$.

Therefore, there exist infinitely many numbers $m$ such that $Q(m) + Q(-m) = 0$, i.e., the polynomial $Q(x) + Q(-x)$ has infinitely many roots. This can occur only when this polynomial is the zero polynomial. That is, we have the identity $Q(x) + Q(-x) = 0$ and so, the graph of $y = Q(x)$ is symmetric about the point $(0; 0)$.

Therefore, the graph of $y = P(x)$ has the center of symmetry at the point $(d; 0)$. We have done.