Show that $x=y+z$ for all $x \in S$

Since $S = \emptyset$ does not satisfy (3), $S$ is nonempty, say $x \in S$. Now (2) implies that $1 = x/x \in S$. It then follows that $S$ is closed under reciprocals, and hence also under multiplication. So it suffices to show that $1$ can be written as $y + z$ for some $y,z \in S$, because then each $x = xy + xz$ is a sum of elements of $S$.

Suppose for contradiction that $1 - y \not \in S$ for every $y \in S$. Since $\{1, -1\}$ does not satisfy (3), we can choose $x \in S$ with $|x| \not= 1$. Since $S$ is closed under multiplication, $x^2 \in S$, and so $1 - x^2 \not\in S$ by our assumption.

Condition (3) implies that $Q \setminus S \subseteq qS \cup \{0\}$. This is actually equality because we are given $0 \not \in S$, and if $x \in S \cap (qS \setminus \{0\})$ then $x = qy$ for some $y \in S$, and then $q = y/x \in S$, which violates (3).

So $1 - x$ and $1 - x^2$ both belong to $qS$, and therefore $\frac{1}{1 + x} = (1 - x)/(1-x^2) \in qS/qS = S/S \subseteq S$. Rewriting $\frac{1}{1+x} = 1 - \frac{x}{1+x}$, we obtain $1 - \frac{x}{1+x} \in S$. Since $\frac{x}{1+x} \in S$ as well, this contradicts our assumption.