We denote by $T(G)$ the set of $2$-tuples $\{(g,n):g\in G,n\in\Bbb N^+\}$ again, but this time we create a free group out of it and quotient by the relation given by $(g,n)^n\sim (e,1)$ as well as $(g,1)(h,1)\sim(gh,1)$ (for all elements $g,h\in G$ and naturals $n$) on top of the equivalence
$(g,n)\sim(h,m)\iff g^m=h^n, $
and write $D(G)=F(T(G))/\sim$. This is a group. If commutativity is absent, we no longer can write products of tuples as other tuples in general. Here $(g,n)$ is an avatar for the formal expression $g^{1/n}$.
But $D(G)$ is not necessarily divisible: products of tuples that can't be simplified don't necessarily have roots. Yet it's still true that $G$ embeds into $D(G)^{[n]}$ as a subgroup for any natural $n$. Moreover, we can form a direct system with the embeddings
$G\hookrightarrow D(G)\hookrightarrow D(D(G))\hookrightarrow D(D(D(G)))\hookrightarrow\cdots.$
Define $\mathrm{Div}(G)$ to be the direct limit of this system. If $x\in \mathrm{Div}(G)$, then $x\in D^m(G)$ for some $m$, and then $v=(x,n)\in D^{m+1}(G)$ satisfies $v^n=(e_{D^m(G)},1)=e$. It follows that $\mathrm{Div}(G)$ is divisble.
One thing I'm struggling with: is there a sense in which $\mathrm{Div}(G)$ is the smallest divisible group containing $G$? Intuitively this seems right, but I'm unsure of how to formally describe/justify this.