Here are two partial answers:

- EDIT: (the following is
*tentative*; in light of Jeremy Rickards example, which *is* vertex-3-connected, and with which I cannot find anything wrong, something must be wrong here; perhaps it is where I leaped to the conclusion that the matroids must be isomorphic) In classical logic, and under the additional hypothesis that both $G$ and $H$ are vertex-3-connected, the answer is no.

One way to prove 1. this is *via infinite matroid theory* by

noting that loc. cit. is a theorem of classical logic,

noting that the contrapositive of the 'Moreover, [...]' in loc. cit. is

If two vertex-3-connected graphs have isomorphic topological cycle matroics, then they are isomorphic.

- noting that the hypothesis of the OP $\color{red}{\text{evidently ensures}}$ ${}^{\text{dubious; I was not thiking carefully here}}$ that the topological cycle matroids are isomorphic, because, so to speak, the matroids 'do not see' putative non-edges-mapped to edges (which the OP's hypotheses at least hypothetically allow).

I do not know whether there are positive examples for what the OP is asking for if both $G$ and $H$ have connectivity 2. (I.e., if both $G$ and $H$ are vertex-2-connected yet both do contain separators of cardinality 2.)

- Even in intuitionistic logic, we have:
**Lemma.** Under the OP's hypotheses, *if* at least one of the maps $f$ and $g$ i *not* a graph-isomorphism, then both $G$ and $H$ are non-forests. (I.e., contain a circuit.)

Proof of the Lemma. Let the data be given as stated. Since the statement is invariant under swapping '$f$' and '$g$', we know that $f$ is not a graph-isomorphism; therefore *by definition* there exists (I think this step is intuitionistically valid (i.v. for short), it is just the *definition* of 'graph-homomorphism $f\colon G\to H$ whose underlying set-map $V(G)\to V(H)$ is injective but which is not a graph-isomorphism) a two-set $xy\in\binom{V(G)}{2}$ with $xy\notin E(G)$ yet $f(x)f(y)\in E(H)$. Since $G$ is a tree, there exists a unique $x$-$y$-path $P_{xy}$ in $G$. Since $xy$ is not an edge of $G$, we know that $P_{x,y}$ has at least two edges (I think this step, too, is i.v.). By hypothesis, $f$ maps (and this is intuitionistically valid: $P_{x,y}$ is finite by definition of 'graph-theoretic path, so the image $f(P_{x,y})$ can be constructed) $P_{x,y}$ to a graph-theoretic path $f(P_{x,y})$ in $H$. Since we assumed that $f(x)f(y)$ is an edge of $H$, we have constructed a circuit $f(x) f(P_{x,y})f(y)f(x)$ in $H$. Now we apply the given injective graph-homomorphism $g$ to said circuit to also construct a circuit in $G$. We have now found circuits in both $G$ and $H$, completing the proof.

**Corollary.** If at least one of the graphs $G$ and $H$ is a *trees*, then the answer to the OP is no, too.

**Corollary.** Under classical logic: if the correct answer to the question in the OP is yes, *then* every example of what the OP is asking for must consist of infinite graphs $G$ and $H$ which both contain at least one separating vertex-set of size 2 and both contain at least one circuit.

4more comments