Are comparability graphs closed under edge contractions?
https://en.wikipedia.org/wiki/Comparability_graph
https://en.wikipedia.org/wiki/Edge_contraction#Vertex_identification
It seems like they are, since any function identifying vertices would correspond to a monotone map between two orientations.
It's unclear whether you're asking about edge contractions or about all vertex identifications (your title and your question disagree) but the answer for both of them is no.
A cycle of length $6$ is a comparability graph, for example for the poset with elements $\{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}$ ordered by inclusion, as shown below.
If you contract any edge, you get a cycle of length $5$, which is not a comparability graph because it's not perfect.
Or, for a self-contained demonstration: suppose that $C_5$ were a comparability graph. Orient each edge $vw$ to point in the direction $v \to w$ when $v < w$. Because the cycle is odd, the edges cannot alternate direction all the way around the cycle: there must be two consecutive edges $uv, vw$ oriented $u \to v, v \to w$ or $u \gets v, v \gets w$. But now, transitivity demands that the edge $uw$ should also exist, and it doesn't: contradiction!