I'll suggest an approach. I'll not focus on $T$ especially: let $H$ be a finitely presentable group, pick a generating set $S$; how can we prove that the unlabeled Cayley graph (denoted $(H,S)$) is isolated among vertex-transitive graphs?

1) a necessary condition for this isolation statement is that $(H,S)$ of $H$ is isolated among labeled Cayley graphs. This condition is simply called, for a group, "isolated", because it does not depend on the generating subset. It is equivalent to be finitely presented and *finitely discriminable*, where the latter means: having a finite subset $\Omega\subset H\smallsetminus\{1\}$ such that every nontrivial subgroup $N$ of $H$ satisfies $N\cap\Omega\neq\emptyset$. Simple groups, finite groups, are finitely discriminable, but there are many others; for instance Thompson's group $F$.

2) Now here is a condition (X) for $(H,S)$ which implies, if $H$ is isolated, that $(H,S)$ is isolated among vertex-transitive graphs:

(X): there exists $R_0\ge 1$ such that every automorphism of the $R_0$-ball in $(H,S)$ fixing 1 is the identity.

Indeed, assume that $(H,S)$ satisfies (X). Let $Y$ be a (non-empty) vertex-transitive graph whose $R_0$-balls are isomorphic to the balls of $(H,S)$. Let $G$ be the automorphism group of $Y$. We claim that $G$ acts freely on $Y$. Indeed, assume otherwise, hence there exists a vertex $v$ and $g\in G$ fixing $v$ but not fixing some neighbor $w$ of $v$. Since $g$ is an automorphism of the $R_0$-ball around $v$ and by (X), we deduce that $g$ fixes $w$, a contradiction.

Now let us be more precise on the labeling: $(H,S)$ has oriented edges labeled by generators (we can assume wlog $1\notin S$). Now let $v$ be a vertex in $Y$ and $e$ an oriented edge in $Y$ (here I just mean an ordered pair of vertices in $Y$ linked by an edge, since $Y$ is not an oriented graph!), contained in the $R_0$-ball around $v$. Then there exists a unique isomorphism $f_v$ from $B_Y(v,R_0)$ to $B_1((G,S),R_0)$ mapping $v$ to 1, let $s(v,e)\in S$ be the label of the image of $e$ by this isomorphism.

Let us show that $s(v,e)$ does not depend on $v$: it is enough to show that $s(v,e)=s(v',e)$ if $v,v'$ are neighboring vertices such that $e$ is in the $R_0$-ball around both. $f_v$ maps $v'$ to some element $t\in S$. Hence, writing $L_g(h)=gh$ in $H$, we see that $L_t^{-1}f_v$ maps $v'$ to 1, and by uniqueness we get $f_{v'}=L_t^{-1}f_v$. $L_t$ preserves the labeling, we deduce that $s(v,e)=s(v',e)$.

Hence with this labeling, $Y$ has the same $R_0$-balls as $(H,S)$.

On the other hand, the definition of the labeling was canonical and therefore it is invariant by isometries of $Y$. Since, after fixing some vertex 1 in $Y$, $Y$ is the Cayley graph of $G$ with respect to its 1-sphere (call it $S'$), this shows that this is the Cayley labeling of $S'$, modulo the canonical bijection $S'\to S$ mapping $s'$ to the label of $(1,s')$.

If now $H$ is isolated and $R_0$ has been chosen large enough, this implies that $S'\to S$ extends to an isomorphism $G\to H$.

So you should check (X). You first need to check that the isometry group of the Cayley graph is reduced to $H$, I don't know if it holds for $T$, and maybe this requires a good choice of generating set to avoid obvious symmetries.

labeledCayley graphs is yes. Indeed all of $F,T,V$ are isolated as marked groups. For $T,V$, it holds because they are finitely presented simple groups. For $F$, it's because it's finitely presented with any proper quotient abelian; that is was isolated was observed in my J. Algebra paper with Guyot and Pitsch, see arxiv.org/abs/math/0511714 $\endgroup$