= Solution
If $X$ is finite and nonempty, let $n=|X|\geq1$, choose a <bijection> with the <cyclic group> $\mathbb Z/n\mathbb Z$, and transport its addition to $X$. This includes the one-element case.
If $X$ is infinite, use the maximal partial <group> $(A,m)$ from the preceding part. Its carrier must be infinite: otherwise $|X\setminus A|<|A|$ would make the complement finite too, forcing $X$ finite. Since the complement injects into $A$,
$$
|A|\leq|X|=|A|+|X\setminus A|\leq|A|+|A|=2|A|=|A|,
$$
where the last equality is the permitted infinite-cardinal doubling identity. Cantor-Schröder-Bernstein supplies a <bijection> $j:X\to A$. Define
$$
\boxed{x\star y=j^{-1}(m(j(x),j(y))).}
$$
<Associativity> follows by applying $j$ and using <associativity> of $m$; the identity is $j^{-1}(e_A)$ and the inverse of $x$ is $j^{-1}(j(x)^{-1})$. Hence $(X,\star)$ is a <group>.
This final transport need not extend the earlier partial operation on its original <subset>. That stronger claim is neither needed nor generally possible: for example a <group> of order three cannot occur as a <subgroup> of a <group> of order four. The proof establishes exactly the existence of some <group structure on every nonempty set>, using the allowed choice consequences in the infinite case.
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