As Benjamin Steinberg says, it does not work in general, and you can see the problem already for 1-dimensional complexes. Suppose you take a Cayley graph for $G$. This is a graph (or 1-dimensional CW-complex) with vertex set $G$ and edge set $G\times S$, where $S\subseteq G$ is a set of elements that generates $G$. The edge $(g,s)$ joins the two vertices $g$ and $gs$. I have set this up so that $G$ acts on the left: for $h\in G$, $h(g,s)=(hg,s)$ defines the action. As we expected, the edge $h(g,s)$ joins the two vertices $hg$ and $hgs$. Since the edge set and the vertex set are both free as $G$-sets, it is possible to make them into $G-G$-bisets. But there isn't usually a way to do this that preserves the incidence relation between vertices and edges. The first thing you might try is to define the right action on edges by $(g,s)k)=(gk,s)$. But then the edge $(g,s)k$ joins the vertices $gk$ and $gks$, whereas the images under $k$ of the ends of the edge $(g,s)$ are $gk$ and $gsk$. If the subset $S$ is closed under conjugation, then you could define $(g,s)k=(gk,k^{-1}sk)$, but if $S$ is not closed under conjugation (which it won't usually be) this won't work. In the case of the universal cover of the wedge of two circles, $S$ is a set of two elements that freely generate the fundamental group.

In terms of $G$-sets, you can think about the vertices and directed edges as free $G$-sets, and the map taking a directed edge to its initial vertex as a $G$-map. The left $G$-maps from $G$ to $G$ are exactly the right multiplications by elements of $G$, and unless $G$ is abelian, many of these will fail to commute with the right $G$-action. So although you can make any free $G$-set into a $G-G$-biset, most left $G$-maps will not be $G-G$-bimaps.

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