[proofplan]
We apply the $3$-connected uniqueness corollary of Whitney's $2$-isomorphism theorem. Equality of the cycle matroids on the same labelled edge set means that the identity map on labels is a matroid isomorphism preserving exactly the simple cycles. Whitney's corollary says that, for finite $3$-connected graphs with at least four vertices, such a labelled cycle-matroid isomorphism is induced by a unique labelled graph isomorphism. Thus the matroid equality produces the required vertex bijection directly, without arguing about intermediate Whitney flip sequences.
[/proofplan]
[step:Interpret equality of cycle matroids as a labelled matroid isomorphism]
Let $E$ denote the common labelled edge set of $G$ and $H$. Let $M(G)$ denote the cycle matroid of $G$ on the ground set $E$: a subset $C \subset E$ is a circuit of $M(G)$ exactly when $C$ is the edge set of a simple cycle in $G$. The equality $M(G)=M(H)$ means that the identity map
\begin{align*}
\operatorname{id}_E:E &\to E
\end{align*}
is an isomorphism from $M(G)$ to $M(H)$ preserving every edge label.
We invoke the $3$-connected uniqueness corollary of Whitney's $2$-isomorphism theorem: if $X$ and $Y$ are finite $3$-connected graphs with at least four vertices and $\psi:E_X\to E_Y$ is an isomorphism from $M(X)$ to $M(Y)$, then $\psi$ is induced by a labelled graph isomorphism $X\to Y$. The hypotheses are satisfied with $X=G$, $Y=H$, and $\psi=\operatorname{id}_E$, because the theorem assumes $G$ and $H$ are finite $3$-connected graphs with at least four vertices and because $M(G)=M(H)$ as matroids on $E$.
[guided]
The cycle matroid records the labelled simple cycles of a graph. Thus the hypothesis
\begin{align*}
M(G)=M(H)
\end{align*}
means more than abstract matroid isomorphism: both matroids have the same ground set $E$, and the identity map on $E$ preserves exactly the circuit subsets. Equivalently, a subset $C\subset E$ is the edge set of a simple cycle in $G$ if and only if the same labelled subset $C$ is the edge set of a simple cycle in $H$.
The needed form of Whitney's theorem is the $3$-connected uniqueness corollary of Whitney's $2$-isomorphism theorem. It states that if $X$ and $Y$ are finite $3$-connected graphs with at least four vertices and $\psi:E_X\to E_Y$ is an isomorphism from the cycle matroid $M(X)$ to the cycle matroid $M(Y)$, then there is a graph isomorphism from $X$ to $Y$ that induces $\psi$ on edge labels.
We verify its hypotheses. The theorem statement gives that $G$ and $H$ are finite $3$-connected graphs with at least four vertices. The equality $M(G)=M(H)$ on the common labelled edge set $E$ says exactly that
\begin{align*}
\operatorname{id}_E:E &\to E
\end{align*}
is an isomorphism from $M(G)$ to $M(H)$. Therefore Whitney's $3$-connected uniqueness corollary applies with $X=G$, $Y=H$, and $\psi=\operatorname{id}_E$.
This direct corollary is the point at which $3$-connectedness is used. We do not need to analyze a sequence of Whitney flips or cut-vertex operations, since such a sequence could pass through intermediate graphs that are not $3$-connected. The corollary packages Whitney's theorem in exactly the endpoint form required here.
[/guided]
[/step]
[step:Record the endpoint consequences of $3$-connectedness]
A cut vertex of a graph $X=(V_X,E_X)$ is a vertex $v\in V_X$ such that the induced graph $X-v$ on $V_X\setminus\{v\}$ is disconnected. Since $G$ is $3$-connected, deleting fewer than three vertices from $G$ cannot disconnect it. In particular, deleting one vertex cannot disconnect $G$, so $G$ has no cut vertex. The same argument applies to $H$.
A $2$-vertex separation of $X$ is a pair of distinct vertices $a,b\in V_X$ such that the induced graph $X-\{a,b\}$ on $V_X\setminus\{a,b\}$ is disconnected. Since $G$ and $H$ are $3$-connected and have at least four vertices, neither graph has a $2$-vertex separation. These endpoint facts are consistent with, but are not used as a substitute for, Whitney's $3$-connected uniqueness corollary.
[/step]
[step:Extract the labelled graph isomorphism]
By Whitney's $3$-connected uniqueness corollary applied to the matroid isomorphism $\operatorname{id}_E:E\to E$, there exists a graph isomorphism
\begin{align*}
\varphi:V_G &\to V_H
\end{align*}
that induces $\operatorname{id}_E$ on the labelled edge set. Therefore, for every labelled edge $e\in E$ with endpoints $u,v\in V_G$, the same labelled edge $e$ has endpoints $\varphi(u),\varphi(v)\in V_H$. This is precisely an isomorphism of $G$ and $H$ as labelled graphs.
[/step]
