Solution:
Set $u = x - \frac{2}{x}$. Then the equation becomes
$$u^{2} + 2u + a^{2} - a - 5 = 0$$
Since the equation $x^{2} - u x - 2 = 0$ has real solutions for any real $u$, it suffices to find the integer values of $a$ for which the equation (1) has a real root. The last holds when $D = -a^{2} + a + 6 \geq 0$, i.e. $(a - 3)(a + 2) \leq 0$ giving $a \in [-2, 3]$. Therefore $a = -2, -1, 0, 1, 2, 3$.