In terms of $G(x) := P(x) - \frac{c}{2}$, the given condition can be rewritten as
$$(x+1) G(x-1) = (x-1) G(x) \text{ for all } x \in \mathbb{R} .$$
It follows immediately (with $x = \pm 1$) that $G(0) = 0$ and $G(-1) = 0$. Therefore, $G(x) = x(x+1) Q(x)$ for some $Q \in \mathbb{R}[x]$. Then
$$Q(x-1) = Q(x) \text{ for all } x \in \mathbb{R} \setminus \{-1, 0, 1\} .$$
This is true iff $Q$ is a constant; i.e.,
$$P(x) = k x(x+1) + \frac{c}{2} \quad \forall x \in \mathbb{R}$$
where $k$ is a real constant.

Remark. We have $(x+2) G(x) = x G(x+1)$ for all $x \in \mathbb{R}$. This is a special case of Problem 2 in the test for Level 4+ (where $a = -1, b = -2$ ).