Answer: $-2 < a < 0$ and $0 < a < 2$.

Letting $c = x + y + z$, one sees that $x, y, z$ are the roots of the polynomial $P(t) = t^3 - 3t + a + 3c$. Therefore, Vieta's theorem implies $x + y + z = 0$, thus $c = 0$ and $P(t) = t^3 - 3t + a$. Consequently the polynomial $P(t) = t^3 - 3t + a$ has three distinct real roots. Moreover, the denominators appearing in the problem statement must be nonzero, hence the roots of $P(t)$ are nonzero. On the other hand, the converse statement also holds: if the polynomial $P(t) = t^3 - 3t + a$ has three distinct nonzero real roots, the equalities

in the problem statement are satisfied by these roots $x, y, z$.
Now for $a > 2$, one would have $t^3 + a > t^3 + 2 \geq 3t$ for positive $t$, thus the polynomial $P(t)$ would have no positive root, so it could not have three real roots. Similarly for $a < -2$, there would be no negative root and there could not be three real roots. Moreover for $a = 2$ and $a = -2$, the polynomial $P(t)$ would have a double root of $t = 1$ and $t = -1$, respectively. Finally for $a = 0$, one of the roots of $P(t)$ would be 0.
Outside all these cases, one has $-2 < a < 0$ or $0 < a < 2$. Indeed for $0 < a < 2$, one has $P(0) = a > 0$ and $P(1) = a - 2 < 0$, hence there is a root in each of the intervals $(-\infty, 0)$, $(0, 1)$ and $(1, \infty)$. Similarly for $-2 < a < 0$, there is a root in each of the intervals $(-\infty, -1)$, $(-1, 0)$ and $(0, \infty)$.