Question: For a given rational number $r\in (0,1)$, does there exists a finite $S\subset \mathbb{N}$ such that every pair of elements of $S$ are coprime and
$$r-\sum_{n\in S}\frac{1}{n}=\frac{1}{b}$$
for some $b\in\mathbb{N}$?
Example: For example, for $r=\frac{3}{11}$ we have
$$\frac{3}{11}-\frac{1}{5}-\frac{1}{14}=\frac{1}{770}$$
For $r=\frac{3}{13}$ we have
$$\frac{3}{13}-\frac{1}{5}-\frac{1}{33}=\frac{1}{2145}$$
For $r=\frac{4}{13}$ we have
$$\frac{4}{13}-\frac{1}{4}-\frac{1}{19}-\frac{1}{199}-\frac{1}{28089}-\frac{1}{502057679}=\frac{1}{2772681042969479772}$$
Work so far: I've verified that this works for all fractions with denominator less than or equal to $13$ using a combination of a greedy algorithm as well as brute force. The greedy algorithm is described below
Step 1: Check whether $r=\frac{1}{b}$ for some $b\in\mathbb{N}$. If yes, set $S=\emptyset$ and you are done. If not go to step $2$.
Step 2: Define $S_1=\{\lceil r^{-1} \rceil\}$ and go to step 3 while setting $i=1$.
Step 3: Check if
$$r-\sum_{n\in S_i}\frac{1}{n}=\frac{1}{b}$$
for some $b\in\mathbb{N}$. If yes, set $S=S_i$. If not go to step 4.
Step 4: Consider
$$m= \left\lceil \left(r-\sum_{n\in S_i}\frac{1}{n}\right)^{-1}\right\rceil$$
and define $S_{i+1}$ to be
$$S_{i+1}=S_i\cup\{\text{Least }n\geq m: \gcd(n,a)=1\text{ for all }a\in S_i\}$$
Go back to step 3 with $i$ incremented to $i+1$.
I actually believe that this greedy algorithm will always end since the numerators of
$$r-\sum_{n\in S_i}\frac{1}{n}$$
seem to 'stay close' to $1$, but unfortunately the sizes of the denominators grow too fast to easily use it to calculate sets $S$ for a significant number of possible $r$.
Motivation: If answered in the affirmative, then this would settle a conjecture I asked about here. The conjecture is: Let $r$ be a given positive rational and let $T$ be a finite set of rational numbers in $(0,r)$. Does there exist a finite set $S\subset \mathbb{N}$ such that
$$\sum_{n\in S}\frac{1}{n}=r$$
but
$$\sum_{n\in R}\frac{1}{n}\not\in T$$
for all $R\in P(S)$ (the powerset of $S$)?
EDIT: As the bounty is over halfway done, I'll add this addendum which might be easier to solve. A well known algorithm for egyption fractions is the greedy algorithm. In terms of the algorithm presented above (in step 4), it would define
$$S_{i+1}=S_i\cup\{m\}=S_i\cup\left\{\left\lceil \left(r-\sum_{n\in S_i}\frac{1}{n}\right)^{-1}\right\rceil\right\}$$
That is, simply pick the smallest integer such that you don't overshoot the rational number you are trying to reach. This algorithm is well known to terminate in all cases. In a similar manner, we could instead modify the algorithm above (again in step 4) to be
$$S_{i+1}=S_i\cup \{\text{Least }k\geq m:k\not|\prod_{n\in S_i}n\}$$
As you can see, this allows for many more terms to be added to $S$ as it includes all possible relatively prime numbers. This alternative algorithm also has the added benefit that it is super easy for a computer to test. With it, I checked it terminated at every rational between $0$ and $1$ with denominator less than or equal to $1000$.
