This is not a complete answer to your question, but more an argument which should demonstrate that the property of being quotient-dense seems to have very few to do with the overall density of $S$.
Especially, I found two counterexamples to my former conjecture (see comments to your question) that $S=\{s_n\mid n\in\Bbb N\}$ is quotient-dense if and only if the always increasing sequence $s_n$ is sub-exponential. One of them is another counterexample to your own conjecture, but of another kind as the one given in the comments.
$s_n$ is sub-exponential, but $S$ is not quotient-dense
This not quotient dense set $S$ is generated by a sequence $s_n<7n$, obviously sub-exponential. To write down the set nicely, I will write $[a,b]_\Bbb N$ for the range of natural numbers from $a$ to $b$, e.g. $[1,6]_\Bbb N$ is short for $1,2,3,4,5,6$. Now we can define
$$S=\{[a_0,b_0]_\Bbb N,\;[a_1,b_1]_\Bbb N,\;[a_2,b_2]_\Bbb N,\;...\}$$
with $a_0=b_0=1$ and recursively $a_i=3b_{i-1}$ and $b_i=\lfloor 3/2\cdot a_i \rfloor$. The following picture shows the generating sequence $s_n$ up to $n\approx 80$ together with the bounding linear function (in gray).
It remains to show that this sequence is bounded by $7n$ and $S$ is not quotient-dense. The former statement is not hard to show but it is annoying to do nicely (I tried). So, I will only show the latter one:
Note that $a_0=b_0<a_1<b_1<a_2<b_2<\cdots$. We will see that no rational number $x\in(1/3,2/3)$ can be expressed by quotients $p,q\in S$. Lets say $p\in[a_i,b_i]_\Bbb N$ and $q\in[a_j,b_j]_\Bbb N$. Because we want to build a rational $<1$, we can assume $i\leq j$. At first assume $i<j$. Then
$$\frac pq\leq\frac {b_i}{a_j}\leq\frac{b_{j-1}}{a_j}=\frac{b_{j-1}}{3b_{j-1}}=\frac13.$$
In the other case $i=j$ we have
$$\frac pq\geq\frac {a_i}{b_i}=\frac {a_i}{\lfloor 3/2\cdot a_i\rfloor}\geq\frac {a_i}{3/2\cdot a_i}=\frac23.$$
So we indeed have this gap $(1/3, 2/3)$ that we cannot fill.
$s_n$ is growing exponentially, but $S$ is quotient-dense
At first note that it suffices to show that any number in $[0,1]$ can be approximated by $s_n/s_m$. Because given a number $x>1$, we can instead approximate $1/x\in(0,1)$ by a sequence $s_{n(i)}/s_{m(i)}$ and then convert this via exchanging nominator and denominator to the sequence $s_{m(i)}/s_{n(i)}$ which converges to $x$.
Choose an enumeration $p_i/q_i,i=1,2,3,...$ of all rational numbers in $(0,1)$. If all of these numbers are representable from $S$, then all of $[0,1]$ can be approximated. Now choose $s_1=1$ (we skip $s_0$ for no real reason) and define recursively
$$s_{2i}=2p_i \cdot s_{2i-1},\qquad s_{2i+1}=2q_i \cdot s_{2i-1}.$$
Note that because $p_i<q_i$, we indeed have $s_{2i}<s_{2i+1}$. Now we have $s_{2i}/s_{2i+1}=p_i/q_i$, hence every rational numbers in $(0,1)$ can be generated from this sequence. We have to show that the sequence is growing exponentially, but this is easy:
$$s_{2i+1}=2q_i\cdot s_{2i-1}\geq 2s_{2i-1}$$
which is an exponentially growing sub-sequence. And because $s_{2i+1}=s_{n(i)}$ with linearly growing $n(i)$, this suffices to imply that $s_n$ itself grows exponentially.
Moral of the story
If you want a nice characterization of quotient-dense sets, you should not look at properties like density. This involves your conjecture on $\sum_n s_n<\infty$ and my conjecture about sub-exponential growth.
Update
Actually any arbitrarily sparse set $S=\{s_n\mid n\in\Bbb N\}$ can be made quotient-dense by at most "doubling its density". Take an enumeration of the rationals $p_n/q_n$ as above. Then we have this new quotent-dense set
$$S':=\{p_ns_n\mid n\in\Bbb N\}\cup\{q_n s_n\mid n\in\Bbb N\}.$$
This new set has "dense" clusters of two elements and big gaps between these clusters. So it seems to have more to do with local clustering than global structure.
