Here is a construction. Every finite group with $n$ elements embeds into $A_{2n}$ (and even $A_{n+2}$) which is simple if $n>2$ and of commutator width 1 (as any other finite simple group by the Ore conjecture proved by Martin W. Liebeck, E. A. O'Brien, Aner Shalev, Pham Huu Tiep, although for $A_n$ it was probably known to Frobenius and certainly to Ore in 1951). Suppose that the group $G$ is infinite. Using HNN extensions embed your group into a group $S$ where all pairs of elements of the same order are conjugate (this is the standard application of HNN extensions: use HNN extensions with cyclic associated subgroups to make $G<G_1$ such that all pairs of elements of $G$ of the same order are conjugate in $G_1$; then do the same to $G_1$ and obtain $G_2$,then $G_3,G_4,...$; the union $\cup G_i$ is the desired group $S$). The group $S$ is always simple because the HNN extension with free letter commuting with $1$ in the HNN construction gives a free product with $\mathbb Z$. Hence the normal subgroup generated by any non-trivial element of $S$ has an element of infinite order, hence all elements of infinite orders (see below), hence coincides with $S$.

Take an element $g\in S$. Then among the HNN extensions used to build $S$ there is one with free letter $t$ such that $tgt^{-1}=g$. That is, $tg=gt$. Then $gt$ and $t$ have infinite order in $S$. So there exists $h\in S$ such that $gt=hth^{-1}$. Therefore $g=[h,t]$, and $g$ is a commutator. This works for groups of any infinite cardinality. If your group is infinite, $S$ will have the same cardinality as your group.

Locally finite groupsby Kegel and Wehrfritz, 1973, Remarks p.115. This follows a detailed proof that in an infinite simple group, every countable subset is contained in an infinite countable simple subgroup (Theorem 4.4 therein). In the same remark they also mention that this is a part of general result of universal algebra. $\endgroup$ – YCor Mar 3 at 9:03