You are right, if $\Delta(G)>2$ and all cycles of $G$ are odd then $\chi'(G)=\Delta(G)$.
You can prove this by induction on the number of cycles in $G$. We my assume $G$ is connected, since if it is true for every connected graph we can just colour components separately.
If $G$ has no cycles then it is a tree. Root it at any vertex, and colour edges one by one in order of distance from the root. We can do this using a greedy algorithm with $\Delta$ colours: when we colour an edge, the only incident edges we have previously coloured all meet it at the same endpoint, so there are at most $\Delta-1$ forbidden colours.
If there is exactly one cycle, then we can do the same thing. First, colour the cycle with $3\leq \Delta$ colours. Now colour the other edges in order of distance from the cycle; the same argument works.
If there are two or more cycles, choose two and call them $C_1,C_2$. If they have a vertex $v$ in common, note that there can be no path between the cycles that does not go through $v$, since if there is such a path $P$ we could construct a cycle by going along $P$, round $C_2$ to $v$, and round $C_1$ to the start of $P$. Since both cycles are odd, and we can choose which direction to go round them, we can make this new cycle of either parity, a contradiction. Thus $v$ is a cutvertex, and we can find two graphs $G_1,G_2$, with no common edges and no common vertices other than $v$, such that $G$ is obtained by gluing $G_1$ and $G_2$ together at $v$, and each containing one of the cycles. By induction, we can define two colourings $c_1,c_2$ of $G_1,G_2$ respectively, each with colours from $\{1,...,\Delta(G)\}$. Since $\Delta(G)\geq d_G(v)$ we can reorder the colours for $c_2$, if necessary, so that the set of colours used at $v$ by $c_2$ is disjoint from those used at $v$ by $c_1$.
If $C_1,C_2$ do not have a vertex in common, then by a similar argument there cannot be two vertex-disjoint paths between them (otherwise there would be cycles of either parity using these paths and part of $C_1,C_2$). This means, via Menger's theorem, that there is a single vertex $v$ such that all paths between them go through $v$, and now you can do the same thing.
Best Answer
I think your generalization from the $G \square K_2$ case is the right idea: Whenever a vertex $u \in G$ is missing a color at one of its edges (say the color $i$), it's missing the same color in the other copy of $G$ (in the $G \square K_2$ case) and hence that edge formed by the Cartesian product with $K_2$ may be colored by $i$ without creating a conflict. A similar technique works in the more general case. First, let's see what we're aiming for for $\chi'(G \square H)$:
Note that for a vertex $(u, v) \in G \square H$, we have $deg_{G \square H}(u, v) = deg_{G}(u) + deg_{H}(v)$. Hence by picking $u \in G, v \in H$ each of maximum degree, we see that $\Delta(G \square H) = \Delta(G) + \Delta(H)$. Our job is therefore to construct a proper $(\Delta(G) + \Delta(H))$-edge-coloring of $G \square H$.
For each copy of $G$ in $G \square H$, use an identical $\Delta(G) + 1$-coloring (which exists by Vizing's theorem (as you mentioned in the proof of $\chi'(G\square K_2)=\Delta(G\square K_2)$)). With $\Delta(G) + 1$ colors allowed, some color $i$ is missing from all edges incident to all copies of the vertex $u \in V(G)$ (for any $u$). To color the copy of $H$ whose first coordinates are all $u$, we need only $\Delta(H)$ colors (since $H$ is class 1); we choose to use $i$ along with $\Delta(H) - 1$ additional colors. Doing this for each $u \in G$ yields an edge-coloring of $G \square H$ in $(\Delta(G) + 1) + (\Delta(H) - 1) = \Delta(G) + \Delta(H)$ colors, as needed; this completes the proof.
Remark: I see that you're an experienced user, so I skimped on the details a bit. Let me know if anything needs more filling in.