Your argument is intuitively correct, but formalizing it requires a very careful choice of "smaller than". In particular all of the infinities in question are countable, so in some sense they are the "same size".
But there are two approaches that we can use.
The first is to point out that in real life it is more common to encounter small numbers than big ones. So if you choose a probability distribution for what denominators you expect in practice, and sum up the values of all of the exactly represented numbers, you can get to a fixed probability of even division. And it won't be zero.
Unfortunately your choice of the relative probability of having a 2 in the denominator versus a 3 is extremely arbitrary. So while the approach makes sense, the numbers you get really will be made up. (Though no matter how you make them up, your general statement is guaranteed correct.)
The second approach is to just look at how the count of exactly representable numbers below N scales with both N and your set of primes. This has been studied. The exact counts are a mess, but if you have a set S of prime divisors of size k, then the count is (1+o(1)) * 1/k! * [product over p in S of (log(N)/log(p)].
The upshot is that for large N, more prime factors always wins. And for specific prime factors, you just look at the ratio of logs.
For instance with this approach we can say that log(3)/log(5) = 1.46497352071793... times as many denominators are exactly representable in base 12 as in base 10. (The log(2) factors cancel out.) Therefore base 12 is better.
Of course the second approach says that base 6 is as good as base 12, and it is smaller, so why not use it? Well, the reason why we keep on winding up with base 12 in practical situations is that divisibility by 4 comes up a lot. It isn't just exactly representable that matters, efficiency matters.