One can prove that $(\log x)S(x)$ does tend to a limit, but I don't think the limit is necessarily a nice constant.  Put $M(z) = \sum_{n\le z} \mu(n)/n$.  Note that by the prime number theorem $M(z) \ll (\log z)^{-A}$ for any $A>0$ and $z$ large enough.  I'll prove that 
$$ 
S(x) \sim \frac{1}{\log x} \sum_{k=1}^{\infty} M(k) \log \Big(1+\frac 1k\Big),
$$ 
and the bound on $M(k)$ guarantees that the sum above is convergent.  This constant can of course be computed (and should be checked with your numerics) but I don't see why it should be anything nice.  

To prove the asymptotic formula, note that $S(x)$ counts all $n\le x$ except for those $n$ having a prime factor $p$ larger than $\sqrt{x}$.  Thus 
$$ 
S(x) = M(x) - \sum_{p>\sqrt{x}} \sum_{\substack{n\le x \\ p|n }} \frac{\mu(n)}{n}.  
$$   
The term $M(x) = O(1/(\log x)^2)$ and so we focus just on the second term above.  Writing $n=mp$ this is 
$$
\sum_{p>\sqrt{x}} \frac{1}{p } \sum_{m\le x/p} \frac{\mu(m)}{m}.
$$ 
Now let $1\le k\le \sqrt{x}$ and group the primes $p$ above according to the ranges $x/(k+1) < p \le x/k$.  Thus the sum above equals
$$
\sum_{1\le k\le \sqrt{x}} M(k) \sum_{x/(k+1) <p\le x/k} \frac{1}{p}.
$$ 
Now use the asymptotics for the sum of the reciprocals of primes to see that the inner sum over $p$ above is 
$$ 
\sim \log \frac{\log (x/k)}{\log (x/(k+1))} \sim \frac{\log (1+1/k)}{\log x}. 
$$
From this the desired asymptotic follows.  

The argument above is a quick sketch, and some details would need to be filled in -- but nothing too hard.  The proof follows ideas of [Dress, Iwaniec and Tenenbaum][1] and see that paper for further details in a related calculation. 

[1]: https://eudml.org/doc/152520