First, if $z = \frac{1}{y}$, then the given system becomes
$$\left\{\begin{array}{c} x \cdot y \cdot \frac{1}{y} > 1 \\ x + y + \frac{1}{y} > \frac{1}{x} + \frac{1}{y} + y \end{array} \Leftrightarrow x > 1 .\right.$$
So, all triples $\left(x, y, \frac{1}{y}\right)$, with $x > 1$ and $y \neq 0$, satisfy the given system of inequalities.

Next, note that the range of $f(y) := y + \frac{1}{y}$ is $(-\infty, -2] \cup [2, +\infty)$ and that, for such triples, $S = x + f(y)$. Then for each value $a \in \mathbb{R}$, we need only consider the following cases:

1. If $a > -1$, we can choose $x = a + 2 (> 1), y = -1$, and get $S = a + 2 + f(-1) = a$.

2. If $a < -1$, we let $x = -a (> 1)$ and choose $y \neq 0$ such that $f(y) = 2a (\in (-\infty, -2])$. Then $S = -a + f(y) = a$.

3. If $a = -1$, we choose $x = 1.5 (> 1), y = -2$ and get $S = 1.5 + f(-2) = -1 = a$.

Therefore, $S$ can take on any real values when $x, y, z$ vary.