Let $T=\mathrm{Tor}(G)$ be the torsion subset of $G$, i.e. $$T=\mathrm{Tor}(G):=\{g\in G: g^n=1 \text{ for some } n\geq 1\}.$$
In general $T$ is not a subgroup of $G$, so it doesn't make sense to talk about the "index" of $T$ in $G$. But we can still ask if finitely many translates can cover $G$.
A set $S\subseteq G$ is called syndetic if finitely many translates of $S$ can cover $G$, i.e. $$G=g_1S\cup\cdots \cup g_n S$$ for some $g_1,\ldots,g_n\in G$. If $S$ is a subgroup then this is equivalent to $[G:S]<\infty$.
Of course if $G$ is a torsion group then $G=T$ so of course $T$ is syndetic. Are there are any other ways that $G$ can be a finite union of translates of $T$?
Problem. Let $T:=\mathrm{Tor}(G)$ be the torsion subset of $G$. Suppose that $T$ is syndetic in $G$. Prove that $G$ is a torsion group.
This checks out when $G$ is finitely-generated abelian, because then $T$ is a direct summand of $G$ with $G=\mathbb{Z}^r\oplus T$. So the only way $[G:T]<\infty$ is if $r=0$ and $G=T$.
What about for nonabelian groups? If $G$ is finitely-generated and the commutator subgroup $G'$ has infinite index, we can bootstrap the f.g. abelian case to solve the problem.
Attempt
Maybe we should try to show that $T$ is a subgroup of $G$. Then $[G:T]<\infty$ which immediately implies that $G$ is a torsion group. It is also clear that the subgroup $\langle T\rangle$ generated by $T$ must have (literal) finite index.
An easy application of the Pigeonhole Principle shows that if $T$ has finite index in $G$, then: for every $g\in G$, there is some $k\in \mathbb{Z}$, $|k|\leq n$ such that $g^k=st$ is a product of two torsion elements $s,t\in T$.
Best Answer
Take the infinite dihedral group $D_{\infty} = \{r,s\mid s^2=1, sr=r^{-1}s\}$. The elements of $D_{\infty}$ are precisely the elements of the form $r^is^j$, with $i$ arbitrary and $j=0$ or $1$. The elements of finite order are precisely the identity, and those of the form $r^is$, which have order $2$. Thus, $T=\{r^is\mid i\in\mathbb{Z}\} \cup \{e\}$.
Since $r^i = s(r^{-i}s)\in sT$, then we have that $D_{\infty}=T\cup sT$. But $D_{\infty}$ is not a torsion group, since $r\in D_{\infty}$ has infinite order.
$D_{\infty}$ can also be realized as the semidirect product $\mathbb{Z}\rtimes\mathbb{Z}_2$, where the action is by inversion.
Note that this is an easy example that shows that the set of torsion elements need not be a subgroup in a nonabelian group (though it generates a characteristic subgroup since as a set it is invariant under automorphisms).