Answer: $P(x) = a x(x-1)(x+1)$, where $a \in \mathbb{R}$.

Set $x = 1$ and $x = -1$ in the initial identity
$$(x-1)P(x+1) - (x+1)P(x-1) = 4P(x). \quad (1)$$
Thus we obtain $-2P(0) = 4P(1)$ and $-2P(0) = 4P(-1)$ respectively. Setting $x = 0$ in (1), we obtain $-P(1) - P(-1) = 4P(0)$, so, taking into account two previous equalities, we have $P(-1) = P(0) = P(1) = 0$.

Therefore $0$, $1$, and $-1$ are zeroes of the polynomial $P(x)$, hence $P(x) = x(x-1)(x+1)Q(x)$, where $Q(x)$ is some polynomial. Setting this representation of $P(x)$ in (1), we obtain
$$(x-1)(x+1)x(x+2)Q(x+1) - (x+1)(x-1)(x-2)xQ(x-1) = \\ = 4x(x-1)(x+1)Q(x),$$
whence
$$(x+2)Q(x+1) - (x-2)Q(x-1) = 4Q(x) \quad \text{for all } x \in \mathbb{R}. \quad (2)$$
Set $x = 2$ in (2), then we obtain $Q(3) = Q(2)$. By induction on $k$, show that $Q(k) = Q(2)$ for all positive integers $k \ge 2$. Indeed, let $Q(k) = Q(2)$ for all $k = 2, 3, \dots, m$, where $m \ge 3$. Then setting $x = m$ in (2), we obtain $(m+2)Q(m+1) - (m-2)Q(m-1) = 4Q(m)$, so, taking into account our assumption, we obtain $(m+2)Q(m+1) = (m+2)Q(2)$, i.e. $Q(m+1) = Q(2)$. It follows that $Q(k) = Q(2)$ for all positive integers $k \ge 2$.

Since the polynomial $Q(x) - Q(2)$ has zero value at infinitely many points, we conclude that this polynomial is equal to zero identically, i.e. $Q(x) = Q(2) = a$ for some real number $a$.

Therefore, the polynomial satisfying (1) needs to have the form $P(x) = a x(x-1)(x+1)$. It is easy to verify that any polynomial of such form (for any real $a$) satisfies (1).