http://mathworld.wolfram.com/EllipticCurveGroupLaw.html?fbclid=IwAR3Dtmv8SS-rJXLGSe3Yp5sRMAyZIN92uCkTxJD5f1MPR2c_cJ4Z3pqLWDY

We claim that the answer is all polynomials $P(x)$ of the form $ax^3 + bx + c,$ where $a, b, c \in \mathbb{R}.$ Firstly, it's easy to verify that these work. We will now show that they are the only solutions. Firstly, by shifting $P$ downwards, we can assume, WLOG, that $P(0) = 0.$ Now, letting $x = 0$ implies that $P$ is odd. By setting $x = 1,$ we can get using odd-ness that $P(y+1)(y-1) = P(y)(y+2) - (2y+1)P(1).$ Therefore, this implies that selecting $P(1)$ and $P(2)$ uniquely determines $P$'s values on the integers, and hence fixes $P$ overall. However, observe that for whatever values we select for $P(1), P(2),$ there is a cubic $ax^3 + bx$ which attains those values (system of equations), and so therefore since $P$ is uniquely determined now and $ax^3 + bx$ works, we know that $P \equiv ax^3 + bx.$ But note that we shifted down so that $P(0) = 0$ in the beginning, so all solutions are $ax^3 + bx + c.$

The colinearity condition is equivalent to $$\frac{P(y)-P(x)}{y-x}=\frac{P(z)-P(x)}{z-x}$$. Taking $x=0$ we get $P(y)+P(-y)=2P(0)$ wich means that all powers of $x$ in $P(x)$ are odd . Taking $y=2x$ we have $z=-3x$ and $5P(x)=4P(2x)+P(-3x)$ considering the greatest power of $x$ in $P(x)$ we get $5x^{2n+1}=2^{2n+3}x^{2n+1}-3^{2n+1}x^{2n+1}$. Clearly $n=0,1$ is a solution and for $n\geq 2$ we can inductively show that $3^{2n+1}>2^{2n+3}$. So, $P(x)$ is at most third degree, and checking $P(x)=ax^3+bx+c$ , is indeed a solution.

Taking $y=z+h$ and $x=-2z-h$ for small $h$ gives $(-2z-h,P(-2z-h)),(z+h,P(z+h)),(z,P(z))$ all colinear. But the line going to these points is the tangent to P at $z$ for small $h$, therefore its equation is $Y=P'(z)(X-z)+P(z)$, then we must have: $$P(-2z)=P'(z)(-3z)+P(z)$$Let $a_d$ the leading coefficient of $P$, we must have $a_d(-2)^d=-3da_d+a_d\iff (-2)^d=-3d+1$. The only solutions are clearly $d=1$ and $d=3$, and using the previous relation we easily find that $P=ax^3+bx+c$