Edge-colouring an $\aleph_1$-chromatic graph so every countable vertex colouring meets all edge colours (Erdős #1176)

Let $G$ be a graph with chromatic number $\aleph_1$ (so its vertices cannot be properly coloured with countably many colours, but can with $\aleph_1$). Is it true that there is a colouring of the edges of $G$ with $\aleph_1$ many colours such that, for every colouring of the vertices of $G$ with countably many colours, some vertex-colour class contains edges of all $\aleph_1$ edge colours? Concretely: does every $\aleph_1$-chromatic graph admit an edge-colouring $c:E(G)\to\aleph_1$ so that whenever $V(G)$ is partitioned into countably many classes, at least one class spans, among its internal edges, every one of the $\aleph_1$ edge colours?

Acceptance. FULLY RESOLVES: either (a) a complete ZFC proof (fully written; machine-checkable preferred where feasible) that every $\aleph_1$-chromatic graph $G$ admits an edge-colouring with $\aleph_1$ colours such that every countable vertex colouring has a class spanning all edge colours; or (b) a proof that the statement is independent of ZFC — a model of ZFC containing an $\aleph_1$-chromatic $G$ for which no such edge-colouring exists, complementing the Hajnal–Komjáth consistency of the positive answer; or (c) a ZFC construction of an $\aleph_1$-chromatic $G$ that provably has no such edge-colouring. ADVANCES (each fully proved): establish the positive statement in ZFC for a structurally restricted class of $\aleph_1$-chromatic graphs (e.g. those with a specified obligatory subgraph, or shift/interval graphs); weaken the hypothesis under which the Hajnal–Komjáth consistency holds, or reprove it from a weaker axiom; or reduce the problem to a partition relation or a set-mapping statement. Deliver the written/formalised proof, the independence proof with its model, or the counterexample graph with the defeating vertex colouring.