In the framework of my answer, I think the hexagon holds since (in the strict case)
$(\tilde\beta_{A,C}\otimes id_B)(id_A\otimes\tilde\beta_{B,C})=
p(i(\tilde\beta_{A,C})\otimes id_{iB})p(...)=
p(\beta_{iA,iC}\otimes id_{iB})p(id_{iA}\otimes\beta_{iB,iC})$ $=p((\beta_{iA,iC}\otimes id_{iB})(id_{iA}\otimes\beta_{iB,iC}))=p(\beta_{iA\otimes iB,iC})=p(\beta_{ip(iA\otimes iB),iC})=p(\beta_{i(A\otimes B),iC})= \tilde\beta_{A\otimes B,C}$.

The problem might be in the axioms of a skeleton. Actually the framework is the following. $C$ is a (small) category and $D$ is a full subcategory such that
$Ob(D)\to Ob(C)\to Ob(C)/iso$ is a bijection (this induces a map $\pi:Ob C\to Ob D$). In that case we don't have $p,i$
as above. What one can do is introduce $\tilde C$ whose objects are pairs $(X,f)$,
where $X\in Ob(C)$ and $f\in Iso_C(X,\pi X)$ and $Hom_{\tilde C}((X,f),(Y,g))=
Hom_{C}(X,Y)$. We then have functors $i:D\to \tilde C$ (taking $X$ to $(X,id_X)$)
and $p:\tilde C\to D$ (taking $X$ to $\pi X$ and $\phi\in Hom_{\tilde C}((X,f),(Y,g))=
Hom_{C}(X,Y)$ to $g\phi f^{-1}$). Then $p\circ i=id_D$, we also have a forgetful
functor $\tilde C\to C$. But I don't see how to lift the braided monoidal strcture of $C$ to $\tilde C$ which makes it
difficult to apply my proposal.