For each $k$-coloring of a graph $G$, there is a complete $k$-partite graph in
which $G$ embeds as a spanning subgraph. Suppose $u$ and $v$ are vertices of $G$ such that in any $k$-colouring, they are assigned different colours. Then they will be adjacent in any complete $k$-partite extension.
(We could say that these vertices are "morally adjacent", but note that
this depends on $k$ implicitly.)
Any uniquely $k$-colourable graph that is not complete $k$-partite is not the intersection of its $k$-partite embeddings.
For another example, consider the graph constructed as follows.
Let $H$, with vertex set $\{1,2,3,4\}$,
be the graph obtained from $K_4$ by deleting the edge $34$. Then in any 3-colouring the vertices $3$ and $4$ get the same colour. Extend $H$ by joining a new vertex $5$ to $4$; call the new graph $H_5$. Then in any 3-colouring, the vertices 3 and 4 get the same colour, and vertices 4 and 5 get different colours. So 3 and 5 get different colours in any 3-colouring, although they are not adjacent. It follows that the intersection of the complete 3-partite embeddings of $G$ contains the edge $35$. (The graph $H_5$ is not uniquely
3-colourable.)
I do not see any way of characterizing the graphs that contain a morally adjacent pair of vertices, but the answer to the question is not "$k$-colourable graphs".