It's not true in general, see (a) below, for elements of order 3 (but true and easy for elements of order 2, see (b)).
Write $G=S_3^I$. Let $G_3$ be the 3-Sylow subgroup in $G$. Fix a 2-Sylow subgroup $S$ in $G_3$ and write $G_2=S^I$.
(a) I claim that for $I$ infinite and $Q=G/(\mathbf{Z}/3\mathbf{Z})^{(I)}$, no subgroup of order $3$ has a complement. Here $N=(\mathbf{Z}/3\mathbf{Z})^{(I)}$ is the subgroup of finitely supported functions.
It's not hard to check that a $(\mathbf{Z}/2\mathbf{Z})^I$-submodule of $(\mathbf{Z}/3\mathbf{Z})^I$ (for the obvious action switching the sign of coordinates, so that the corresponding semidirect product is $G$) has the form $(\mathbf{Z}/2\mathbf{Z})^J$ for some $J\subset I$. In particular, submodules are closed for the product topology. Now let $c$ be any element of $G_3$ with full support. If by contradiction $c$ has a complement $M/N$, then the projection of $M$ on $(\mathbf{Z}/2\mathbf{Z})^I$ is the whole group, and hence $M\cap G_3$ is a $(\mathbf{Z}/2\mathbf{Z})^I$-submodule of $G_3=(\mathbf{Z}/3\mathbf{Z})^I$. Since it contains $N$ it is dense, and hence it equals $M$. So $M=G$, and we have a contradiction.
(b) Write $Q=G/N$; write $Q_3$ and $Q_2$ for the image of $G_3$ and $G_2$ in $Q$. Then $Q_p$ is elementary $p$-abelian, $Q=Q_2Q_3$ with $Q_3$ normal, and $Q_2\cap Q_3=\{1\}$. Hence $Q=Q_3\rtimes Q_2$.
Suppose $c\in Q$ has order $2$; let $d$ be a lift of $c$ in $G$; then $d$ has order $2$ or $6$ and so up to replace $d$ with $d^3$, we can suppose that $d^2=1$; then up to conjugate in $G$, we can suppose that $d\in G_2$ and hence that $c\in Q_2$. Then it's clear that $\langle c\rangle$ has a normal complement (first mod out by $Q_3$ and boil down to the case of an elementary abelian 2-group).