The probability of an event $A$ is given by: $$P(A) = \frac{n(A)}{n}$$ Usually, $n$ is easy to find out, and the only hurdle in finding $P(A)$ is $n(A)$.
However, suppose we are interested in finding $n(A)$. We know $n$ and can find $P(A)$ using some other method (not the formula above), then we can find $$n(A) = nP(A)$$
I'm looking for examples of such counting problems, which can be solved by probability.
The method you cite requires the elements in the outcome set to all have identical probability mass (ie: non-biased weighting) [and, of course, a finite count].
If you do have that, then you can indeed use basic algebra to get $\eta(A)=n~\mathsf P(A)$ .
I cannot think of a case where the finding the count is too much harder than finding the probability. In such a setting, if you can find one you could find the other in much the same way. They just differ by a constant factor, after all.
There is a related notion to your idea, probabilistic proofs of combinatorial identities. Suppose you have mutually exclusive and exhaustive events $E_1,\dots,E_m$, which means that $$ P(E_1)+\dots+P(E_m)=1\tag1 $$ Multiplying both sides by $n$, you get $$ n(E_1)+\dots+n(E_m)=n,\tag2 $$ an identity where all numbers involved are integers. Often, for integral identities where all the terms involved have some combinatorial meaning, it is desirable to give a bijective proof, meaning you find two sets describing the two sides of the equation, and give a bijection between the two sets. I will now give an example where $(1)$ is easy to prove, but $(2)$ is hard to give a bijective proof for. In my mind, probabilistic proofs are the next best thing to bijective proofs.
Imagine an urn which initially contains a single red marble and a single blue marble. You make a series of $n$ draws from this urn, where after each draw, you put back the marble you drew, plus two additional marbles of the same color. Let, $E_k$ be the event that you drew exactly $k$ marbles over the course of these $n$ draws. In my answer, I show that $$ P(E_k)=\frac{\binom{2k}k\binom{2(n-k)}{n-k}}{4^n},\qquad k\in \{0,1,\dots,n\} $$ which leads to the identity $$ \sum_{k=0}^n \binom{2k}k\binom{2(n-k)}{n-k}=4^n\tag3 $$ I cannot really claim that $(3)$ is hard to prove. You can automatically prove $(3)$ with a computer, using the methods of the book A=B by Petkovsek, Wilf and Zeilberger. Also, there is a pretty straightforward generating function proof of $(3)$. But these proofs lack the intuitiveness provided by a bijective proof, and proving $(3)$ bijectively is famously hard.
