Solution:

First, note that $100,000=2^55^5.$

Therefore, any element of $S$ must be of the form $2^a5^b$ with $0 \leq a$ and $b \leq 5.$

Suppose I have distinct $x,y \in S$ with $$x = 2^a5^b,$$$$y=2^c5^d.$$ Then, $$xy = 2^{a+c}5^{b+d}.$$ Thus, $$0 \leq a+c,b+d \leq 10.$$ This means that there are $$(10+1)(10+1)=121$$ divisors. However, there are some divisors that can only exist if $x=y,$ where $$(a,b)=(c,d).$$

If $a+c=0,$ then $a=0,c=0$ must be true.

If $a+c=10,$ then $a=5,c=5$ must be true.

With any other value of $a+c,$ we can have $a \neq c.$

Similar structure holds for $b+d.$ Thus, if $$a+c,b+d \in \{0,10\},$$ then $$(a,b)=(c,d),$$ thus making $x=y.$

This means we have to eliminate $4$ choices, leaving $$121-4=117.$$

Thus, the answer is C .