Since $xyz = 1$,
$$\begin{aligned} x + \frac{1}{y} &= y + \frac{1}{z} \iff xyz + z = y^2z + y \\ \iff 1 - y = -z(1 - y^2) \iff y = 1 \text{ or } z = -\frac{1}{1+y} \end{aligned}$$
If $y = 1$, the system reduces to $xz = 1 \iff z = \frac{1}{x}$ and
$$x + 1 = 1 + \frac{1}{z} = z + \frac{1}{x} \iff x + 1 = 1 + x = \frac{2}{x} \iff x = 1 \text{ or } x = -2$$
If $z = -\frac{1}{1+y}$, then $xyz = 1 \iff x = -\frac{1+y}{y}$ and
$$-\frac{1+y}{y} + \frac{1}{y} = y - (1+y) = -\frac{1}{1+y} - \frac{y}{1+y}$$
(Solutions)
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which is true for every $y \neq 0, -1$ (all three expressions are equal to $-1$).
So all solutions are $(1, 1, 1)$ and $(-\frac{1+t}{t}, t, -\frac{1}{1+t})$ and its cyclic analogous triples. Notice that $t=1$ yields the solution $(-2, 1, -\frac{1}{2})$.