smc

The prime factorization of $100,000$ is $2^5\cdot 5^5$. Thus, we choose two numbers $2^a{5}^b$ and $2^c{5}^d$ where $0\le a,b,c,d\le 5$ and $(a,b)\ne (c,d)$, whose product is $2^{a+c}{5}^{b+d}$, where $0\le a+c\le 10$ and $0\le b+d\le 10$. Notice that this is similar to choosing a divisor of $100,000^2=2^{10}{5}^{10}$, which has $(10+1)(10+1)=121$ divisors. However, some of the divisors of $2^{10}{5}^{10}$ cannot be written as a product of two distinct divisors of $2^5\cdot 5^5$, namely: $1=2^0{5}^0$, $2^{10}{5}^{10}$, $2^{10}$, and $5^{10}$. The last two cannot be written because the maximum factor of $100,000$ containing only $2$s or $5$s (and not both) is only $2^5$ or $5^5$. Since the factors chosen must be distinct, the last two numbers cannot be so written because they would require $2^5\cdot 2^5$ or $5^5\cdot 5^5$. The first two would require $1\cdot 1$ and $2^5{5}^5\cdot 2^5{5}^5$, respectively. This gives $121-4=117$ candidate numbers. It is not too hard to show that every number of the form $2^p{5}^q$, where $0\le p,q\le 10$, and $p,q$ are not both $0$ or $10$, can be written as a product of two distinct elements in $S$. Hence the answer is $\boxed{117}$.