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:
- $f(v,t) = p(v)$ for all $v\in V$, that is, the motion begins at $p$.
- $d_S(f(u,t),f(v,t))=d_S(p(u),p(v))$ 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 $G$.
- $\alpha_t(u,v,w)=\alpha_0(u,v,w)$ 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, 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.]
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...