To start things off, here is a simple observation: the set $S$ is contained in the rational interval $\mathbb{Q}\cap[\frac 12,1]$, the rational numbers $\frac ab$ where $0<a\leq b\leq 2a$.
The reason is that $1/1$ has this form and your transformations preserve the property of being in this interval. If $a\leq b\leq 2a$, then $b/2a$ obeys the requirement, since $b\leq 2a\leq 2b$. And if $a/b$ and $c/d$ obey the requirement, then so does $(a+b)/(c+d)$$(a+c)/(b+d)$, since $(a+c)\leq (b+d)\leq 2a+2c$$a+c\leq b+d\leq 2a+2c=2(a+c)$.