What is the cardinality of the set of all groups (up to isomorphism) of countable order?

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Apologies if this is a known result, I have looked around and could not find it.

There is a pretty vast literature on the number of groups up to isomorphism of order $n$, for any natural number $n$. This paper for instance classifies for which $n$ are there are one, two, or three groups of order $n$. However, I haven't been able to find any results, not even partial, on the cardinality of the number of groups up to isomorphism of order $|\mathbb N|$. Is this known, and if not are there partial results on it? I suspected that the number of abelian groups up to isomorphism would be known (as it is a well-known result for finite groups) but I cannot find the result, if it is published.

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There is at most continuum such groups: every group can be defined by function $\mathbb N \times \mathbb N \to \mathbb N$, and there are only continuum such functions.

There is also at least continuum such groups: for every subset of primes $X$, group $\mathbb Z \oplus \bigoplus\limits_{x \in X} \mathbb Z_x$ is countable, and groups for different $X$ are non-isomorphic: if two sets $X_1$ and $X_2$ differ in prime $p$, then one of corresponding groups includes element of order $p$, and other doesn't.