We can see that $x$ must not divide the product of the other elements of $S$  because if it does,$$\mathrm{gcd}\bigg(x,\prod_{y\in S\setminus\{x\}}y\bigg)=x\geq \frac{x}{k}$$ for every $k\geq 1$ .Erdos has proved that every set $S$ with the above property must have  $\pi(n)$ elements at  most.  

> "Let $a_n$ be a sequence of  positive integers with
> $1<a_1<\cdots<a_n\leq N$ which has the property:  
**(A)** $a_i\nmid \frac{a_1\cdot a_2\cdots a_n}{a_i}$ for every $i=1,...,n$ .Then $n\leq \pi(N)$ holds"    

(As  it was asked) here is the   
**proof:** We will see that if a proper choise of maximum number of elements not exceeding $n$ with the mentioned property exists,then we can construct another (equivalent)  set of elements containing only prime powers.    

Suppose that the *maximum* number of elements we can choose from $\{2,...,n\}$ with the mentioned property is $r\geq \pi(n)+1$.  
It is impossible to have all elements prime powers because by the pigeonhole principle there will be 2 elements $p^a,p^b$ with $a<b$ and $p^a|p^b$ which means that the desired property does not hold for $p^a$.  
So,there must be at least one element that can be written as $x=k\cdot m$ with $\gcd (k,m)=1$.    

If $k$ does not divide the product of the rest elements and so does $m$, then we can pull out $x$ from the set and place $k$ and $m$ into the set ,having a new set with $r+1$ elements with property **(A)** holding true.  
 (of course no other of the elements is equal to $k$ or $m$ because this would again lead to a contradiction)  
But this is a contradiction since $r$ is the maximum number of elements as we assumed.  
  
So,without loss of generality we may assume that $k$ does not divide the product of the rest,but $m$ does.  
This means that we can replace $x$ with $k$ in the set with property **A)** holding true.  
We repeat the argument again until $k$ "drops" to a prime power.  
( which lets us arrive at a contradiction  for prime powers as we already mentioned at the beggining)  
  
(By the way this is not Erdos's proof but one i found some years ago.But i am almost sure Erdos proved this theorem)
  
So, your set must have at most $\pi(n)$ elements.  
On the other hand, if your set contains at least $\pi(n)+1$ elements it could contain 2 powers of the same prime,$p,p^m$ and so the gcd you want would be at least $x=p$ (or $x=p^m$).  
I think it would be more difficult to determine the asymptotic size of $S$ depending on $k$ but certainly  $|S|\leq\pi(n)$ holds for a random set $S$ as you require.