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. Edit The limit is in fact $1$ as noted by Terry Tao and user45947. 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.  Edit In fact one can see that $$ \sum_{k=1}^{K} M(k) \log \Big( 1+ \frac 1k \Big) = \sum_{n\le K} \frac{\mu(n)}{n}\sum_{k=n}^{K}\log (1+1/k) = \sum_{n\le K} \frac{\mu(n)}{n} \log \frac{K+1}{n}. $$ Since $\sum_{n=1}^{K} \mu(n)/n = O((\log K)^{-A})$ and $-\sum_{n=1}^{\infty} \mu(n)(\log n)/n=1$, it follows that the constant is in fact $1$.

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 and see that paper for further details in a related calculation.

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. [Edit The limit is in fact $1$ as noted by Terry Tao and user45947. ] 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. [ Edit In fact one can see that $$ \sum_{k=1}^{K} M(k) \log \Big( 1+ \frac 1k \Big) = \sum_{n\le K} \frac{\mu(n)}{n}\sum_{k=n}^{K}\log (1+1/k) = \sum_{n\le K} \frac{\mu(n)}{n} \log \frac{K+1}{n}. $$ Since $\sum_{n=1}^{K} \mu(n)/n = O((\log K)^{-A})$ and $-\sum_{n=1}^{\infty} \mu(n)(\log n)/n=1$, it follows that the constant is in fact $1$. ]

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 and see that paper for further details in a related calculation.

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 and see that paper for further details in a related calculation.

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. Edit The limit is in fact $1$ as noted by Terry Tao and user45947. 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. Edit In fact one can see that $$ \sum_{k=1}^{K} M(k) \log \Big( 1+ \frac 1k \Big) = \sum_{n\le K} \frac{\mu(n)}{n}\sum_{k=n}^{K}\log (1+1/k) = \sum_{n\le K} \frac{\mu(n)}{n} \log \frac{K+1}{n}. $$ Since $\sum_{n=1}^{K} \mu(n)/n = O((\log K)^{-A})$ and $-\sum_{n=1}^{\infty} \mu(n)(\log n)/n=1$, it follows that the constant is in fact $1$.

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 and see that paper for further details in a related calculation.