[proofplan]
The proof reduces the matroid statement to Whitney's graph-theoretic criterion for connected labelled graphs. The circuits of a labelled cycle matroid are exactly the labelled edge sets of graph cycles, so equality of labelled cycle matroids is equivalent to equality of labelled cycle families. Whitney's criterion then says that, for connected graphs with the same labelled edge set, equality of labelled cycle families is equivalent to being related by Whitney $2$-isomorphisms, namely the usual sequence of vertex identifications, separations, and twists preserving edge labels.
[/proofplan]
[step:Identify the circuits of the two labelled cycle matroids]
Let $\mathcal{P}(E)$ denote the power set of the common labelled edge set $E$. Let $\mathcal{C}_G \subseteq \mathcal{P}(E)$ denote the set of circuits of the labelled matroid $M(G)$, and let $\mathcal{C}_H \subseteq \mathcal{P}(E)$ denote the set of circuits of the labelled matroid $M(H)$. By the definition of the cycle matroid of a graph, a subset $C \subseteq E$ belongs to $\mathcal{C}_G$ if and only if $C$ is the labelled edge set of a cycle in $G$. Likewise, $C \subseteq E$ belongs to $\mathcal{C}_H$ if and only if $C$ is the labelled edge set of a cycle in $H$.
[guided]
We first isolate exactly what information the labelled matroids contain. Define $\mathcal{C}_G \subseteq \mathcal{P}(E)$ to be the circuit set of $M(G)$, and define $\mathcal{C}_H \subseteq \mathcal{P}(E)$ to be the circuit set of $M(H)$. These are families of subsets of the common labelled edge set $E$.
For a graph, its cycle matroid is constructed so that the circuits are precisely the minimal dependent edge sets, and those minimal dependent edge sets are exactly the edge sets of cycles. Since the edge set is labelled, no relabelling is hidden here: the subset $C \subseteq E$ is the actual labelled set of edges. Hence
\begin{align*}
C \in \mathcal{C}_G \iff C \text{ is the labelled edge set of a cycle in } G.
\end{align*}
The same definition applied to $H$ gives
\begin{align*}
C \in \mathcal{C}_H \iff C \text{ is the labelled edge set of a cycle in } H.
\end{align*}
Thus the labelled matroid records exactly the labelled cycle family of the graph.
[/guided]
[/step]
[step:Translate equal labelled matroids into Whitney $2$-isomorphism]
Assume $M(G)=M(H)$ as labelled matroids on $E$. Equality as labelled matroids means that the identity map on the labelled ground set $E$ identifies their independent sets, and hence identifies their circuit families as subsets of $\mathcal{P}(E)$. Therefore $\mathcal{C}_G=\mathcal{C}_H$. By the circuit identification above, for every subset $C \subseteq E$, the set $C$ is the labelled edge set of a cycle in $G$ if and only if it is the labelled edge set of a cycle in $H$.
We now use Whitney's graph-theoretic $2$-isomorphism criterion: for connected graphs with the same labelled edge set, equality of the labelled cycle families is equivalent to being $2$-isomorphic in Whitney's usual sense, meaning obtainable from one another by a finite sequence of Whitney twists and the harmless identifications or separations at cut vertices, all preserving the edge labels. The connectedness hypotheses on $G$ and $H$ are exactly the hypotheses under which this criterion compares a single graph rather than separate connected components. Applying the criterion to the equality of labelled cycle families gives that $G$ and $H$ are $2$-isomorphic.
[guided]
Assume $M(G)=M(H)$ as labelled matroids on the common labelled edge set $E$. Because the labels are fixed, this equality is not an abstract matroid isomorphism with an unspecified bijection of ground sets; it says that the identity map on $E$ preserves the matroid structure. In particular, it preserves circuits, so
\begin{align*}
\mathcal{C}_G = \mathcal{C}_H.
\end{align*}
From the previous step, $\mathcal{C}_G$ is the family of labelled edge sets of cycles in $G$, and $\mathcal{C}_H$ is the family of labelled edge sets of cycles in $H$. Thus the equality of circuit families says exactly that every labelled subset $C \subseteq E$ is the edge set of a cycle in $G$ if and only if the same labelled subset $C$ is the edge set of a cycle in $H$.
This is the point where the nontrivial graph theory enters. We apply Whitney's graph-theoretic $2$-isomorphism criterion. Its hypotheses require connected graphs and a common labelled edge set; these hold by the theorem statement. Its conclusion says that equality of labelled cycle families is equivalent to the existence of a finite sequence of Whitney twists, together with the standard identifications and separations at cut vertices, carrying $G$ to $H$ while preserving the labels of the edges. Therefore the equality $\mathcal{C}_G = \mathcal{C}_H$ implies that $G$ and $H$ are $2$-isomorphic in Whitney's usual graph-theoretic sense.
[/guided]
[/step]
[step:Use Whitney $2$-isomorphism to recover equality of labelled matroids]
Conversely, assume $G$ and $H$ are $2$-isomorphic with the common labelled edge set $E$. By Whitney's graph-theoretic $2$-isomorphism criterion, applied again to the connected labelled graphs $G$ and $H$, the labelled cycle families of $G$ and $H$ agree. Thus, for every subset $C \subseteq E$, the set $C$ is the labelled edge set of a cycle in $G$ if and only if it is the labelled edge set of a cycle in $H$. Using the circuit description of cycle matroids, this says precisely that $\mathcal{C}_G=\mathcal{C}_H$.
A matroid is determined by its circuit family: a subset of the ground set is dependent if and only if it contains a circuit, and it is independent if and only if it contains no circuit. Since $M(G)$ and $M(H)$ have the same ground set $E$ and the same circuits, they have the same independent sets. Hence $M(G)=M(H)$ as labelled matroids.
[/step]
[step:Conclude the equivalence]
The first implication shows that equality of the labelled cycle matroids forces labelled $2$-isomorphism, and the second implication shows that labelled $2$-isomorphism forces equality of the labelled cycle matroids. Therefore $M(G)=M(H)$ as labelled matroids if and only if $G$ and $H$ are $2$-isomorphic.
[/step]
