After analyzing this question I started wondering.
Thoughts. Everyone can give a simple example of a countable group $G, |G| = \aleph_0$ which has uncountable number $2^{\aleph_0}$ subgroups, for instance $G = \bigoplus_{n=1}^{\infty}\mathbb{Z}_2$ but if we are going to assume only non-isomorphic subgroups it turns out that $G$ has only countable number of them. We can modify this example to $G = \bigoplus_{p_n}\mathbb{Z}_{p_n}$ ($p_n$ are primes) and it will satisfy the condition. It is obvious that number of countable groups that have $2^{\aleph_0}$ non-isomorphic subgroups is uncountable since we can construct $2^{\aleph_0}$ of them as follows. Denote $\Pi_1 = \{p_1, p_3, p_5, \dots\}, \Pi_2 = \{p_2,p_4, p_6, \dots\}$. Now our set will be $\{(\bigoplus_{p_k \in \Pi_1}\mathbb{Z}_{p_k})\oplus (\bigoplus_{p'_k \in A}\mathbb{Z}_{p'_k}) \mid\forall A \subset \Pi_2\}$. Obvious that this set is uncountable and all groups from this set are countable and has $2^{\aleph_0}$ non-isomorphic subgroups.
Problem. But what is about cardinality of the set $X$ ($X_A$ for abelian) of all countable groups that have $\aleph_0$ non-isomorphic subgroups?
Solution (for finitely generated abelian case) If we will assume only abelian groups then $|X_A| = \aleph_0$ since we can decompose each group into direct sum of prime ($\mathbb{Z}$ or $\mathbb{Z}_p$) and if number of non-isomorphic summands will be infinite then number of non-isomorphic subgroups is uncountable. Hence every $G \in X_A$ has finite number of non-isomorphic summands. And there are countably many different prime summands. So finally we got that $|X_A| = \aleph_0$. But can situation be changed using all abelian or non-abelian groups?
P.S. Where I'm saying "set of groups" one should understand "set of isomorphism classes".
