Existence of certain graph gadget related to coloring odd hole free graph
One can extract an argument that this cannot work from the paper found by OP in the MO thread. Suppose $G=(V,E)$ is as required, and $c:V\to[k]$ is a $k$-coloring. By the assumption, $c(u)\neq c(v)$. Consider the (bipartite) subgraph $H$ induced by $\{x\in V\ |\ c(x)\in\{c(u),c(v)\}\}$. 

If $u$ and $v$ are in the same connected component of $H$, pick any shortest path in $H$ between $u$ and $v$; it is an induced path in $G$, with colors alternating between $c(u),c(v)$, and must have an odd number of edges because the colors at its ends differ. This contradicts the assumption.

So $u,v$ are in different connected components; but then one can toggle the coloring of one of these components to obtain a coloring $c'$ with $c'(u)=c'(v)$, contradiction.