Let $n \in \mathbb{N}$ and let
$$ n = p_1^{a_1} \cdots p_r^{a_r} $$
be its factorization into primes. My intention is to count how many abelian groups $G$ of order $n$ are there. What follows are my thoughts about this question, which surely are quite rudimentary.
By the structure theorem, we have that: $$ G \simeq \bigoplus_{i = 1}^r \bigoplus_{j = 1}^{m_i} \mathbb{Z}/(p_i^{s_{i,j}}) $$
with $s_{i,1} \leq \dots \leq s_{i,m_i}$ for each $i$ and thus, the cardinals of these two groups must coincide:
$$ n = \prod_{i = 1}^r \prod_{j = 1}^{m_i} p_i^{s_{i,j}} = \prod_{i = 1}^r p_i^{\sum_{j = 1}^{m_i}s_{i,j}} $$
Therefore, by the uniqueness of prime factorization,
$$ s_{i,1} + \dots + s_{i,m_i} = a_i \quad (\forall i\in [r]) $$
The question then turns into a combinatorial one, that is, how many combinations of non-negative, non-decreasing integers are there such that:
$$ s_{i,1} + \dots + s_{i,m_i} = a_i \quad (\forall i\in [r]) $$
Since the problem is independent for each prime, we can tackle each of these independently. Moreover, we can count the amount of tuples of non negative numbers $(s_1, \dots, s_{a_i})$ in increasing order such that $\sum_{j = 1}^{a_i}s_j = a_i$; that is, we can fix a size for the amount of numbers, since the smallest possible case is to have exactly $a_i$ ones. Having done this, we have as many combinations as the $a_i$-th coefficient of the following generating function
$$ f_i = \prod_{j = 1}^{a_i}(1-X^j)^{-1} = \left(\sum_{k \geq 0}X^k\right) \dots \left(\sum_{k \geq 0}X^{ka_i}\right) $$
by corresponding solutions to $x_1 + 2x_2 + \dots + a_ix_{a_i} = a_i$ with tuples by taking:
$$ s_{a_i-q} = \sum_{k = q}^{a_i}x_k $$
that is, we would have $[f_i]_{a_i}$ possible solutions, giving a total of
$$ [f_1]_{a_1} \cdots [f_r]_{a_r} $$
possible abelian groups of order $n$.
I'd really appreciate if you could comment on whether this approach is correct or not and if so, how could one get a more explicit answer, since this one is rather unsatisfactory.
The number of Abelian groups of order $p^n$ is the $n$-th partition number $p(n)$. Therefore the number of Abelian groups of order $p_1^{a_1}\cdots p_r^{a_r}$ is $p(a_1)\cdots p(a_r)$.