Here's one approach to formalizing "moved around". Let $G=(V,E)$ be a graph. Let $p:V\rightarrow S$ be a *placement* of $G$, that is, a map from the vertex set of $G$ to $S$. Let us call the data $(G,p)$ a **framework** on $S$. Let us define a *motion* of $(G,p)$ to be a continuous family of placements $f:V\times[0,1]\rightarrow S$ such that:

1. $f(v,0) = p(v)$ for all $v\in V$, that is, the motion begins at $p$.
2. $d_S(f(u,t),f(v,t))=d_S(p(u),p(v))$ for all $uv\in E$ and all $t\in[0,1]$, where $d_S(\cdot,\cdot)$ is the distance function on $S$. This condition just states that the motion preserves the lengths of all edges of $G$.
3. $\alpha_t(u,v,w)=\alpha_0(u,v,w)$ for all triples of vertices $uvw$ such that $uv\in E$ and $vw\in E$, where $\alpha_t(u,v,w)$ is the angle between the geodesic segments $uv$ and $vw$ at $v$. This condition ensures that the motion preserves all angles between adjacent pairs of edges of $G$.

[Such frameworks are related to [the point-line frameworks of Jackson and Owen](https://arxiv.org/abs/1407.4675), and also work of Tay, Whiteley, Jackson and Jordán and others on 2D molecular graphs and frameworks (see e.g. [this paper of Jackson and Jordán](https://doi.org/10.1007/s00454-008-9100-z).]

Your question is essentially: Let $(G,p)$ be the molecular framework constructed from a geodesic triangle $T$ on $S$. Suppose there exists a motion from $(G,p)$ to any other congruent geodesic triangle (i.e. one with the same lengths, angles and orientation as $T$). Does this imply that $S$ has constant curvature?

I suspect the answer is no, for a possibly silly reason. It's not obvious to me that generic embedded surfaces need to have any pairs of distinct congruent geodesic triangles at all; the condition would be vacuous on such surfaces, which also don't have constant curvature. If we add an additional condition on $S$ that such pairs exist, then the answer seems like it might be yes, but I'm not up for proving it at this moment...