Write $r = \frac{a}{b}$ in lowest terms; how $\frac{55a}{b}$ reduces is governed by $\gcd(b, 55),$ which is $1,$ $5,$ $11,$ or $55$

The cases $\gcd = 5$ and $\gcd = 11$ give $2b = 25a$ and $5b = 22a;$ coprimality pins down exactly one fraction in each case

Write $r = \frac{a}{b}$ in lowest terms and let $g = \gcd(b, 55).$ Then $55r$ in lowest terms is $\frac{\frac{55a}{g}}{\frac{b}{g}}$ (no further cancellation is possible since $\gcd(a, b) = 1$ and $55$ is squarefree). The condition is $$a + b = \frac{55a}{g} + \frac{b}{g}.$$ If $g = 1$ this forces $a = 55a,$ impossible; if $g = 55$ it forces $b = \frac{b}{55},$ impossible.

If $g = 5:$ $a + b = 11a + \frac{b}{5}$ gives $\frac{4b}{5} = 10a,$ so $2b = 25a.$ Since $\gcd(a, b) = 1,$ we need $a = 2$ and $b = 25,$ so $r = \frac{2}{25}$ (indeed $2 + 25 = 27 = 22 + 5$ from $55r = \frac{22}{5}$). If $g = 11:$ $a + b = 5a + \frac{b}{11}$ gives $\frac{10b}{11} = 4a,$ so $5b = 22a,$ forcing $a = 5,$ $b = 22$ and $r = \frac{5}{22}$ (with $5 + 22 = 27 = 25 + 2$ from $55r = \frac{25}{2}$).

Hence $S = \left\{\frac{2}{25}, \frac{5}{22}\right\}$ and the sum is $\frac{2}{25} + \frac{5}{22} = \frac{44 + 125}{550} = \frac{169}{550},$ already in lowest terms. The answer is $169 + 550 = 719.$