Suppose that for any two triangles, (pairwise) equal edge sizes imply (pairwise) equal angles between geodesics. Then the curvature has to be constant.

By contradiction, suppose there exist $\epsilon > 0$, $k_1, k_2$ such that $k_1 < k_2 - 3\epsilon$ and points $p_1, p_2$ such that in the (geodesic) ball $B_\epsilon(p_1)$ Gauss curvature is $\le k_1 + \epsilon$ and in $B_\epsilon(p_2)$ Gauss curvature is $ \ge k_2 - \epsilon$. This implies that in $B_\epsilon(p_1)$ all non-degenerate triangles have all angles smaller than the corresponding angles of the comparison triangle in $M_{k_1 + \epsilon}$, where $M_{k_1 + \epsilon}$ is a surface of a constant curvature $k_1 + \epsilon$, and "the comparison triangle" means the triangle with same side lengths. Similarly in $B_\epsilon(p_2)$ all non-degenerate triangles have all angles bigger than the corresponding angles of the comparison triangle in $M_{k_2 - \epsilon}$. 

Now if you take any triangle $T$ with the diameter $<\epsilon/10$ in our surface then you can find triangles with the same lengths as $T$ in both $B_\epsilon(p_1)$ and $B_\epsilon(p_2)$. And since we have that "equal edge sizes imply equal angles"
then we have: 
[the angle in $T$] $=$[the angle in comparison triangle in $B_\epsilon(p_1)]$ $<$ 
 [the angle in comparison triangle in $M_{k_1 + \epsilon}]$ $<$  
 [the angle in comparison triangle in $M_{k_2 - \epsilon}$] $<$
[the angle in comparison triangle in $B_\epsilon(p_2)$] $<$ [the angle in $T$]. And that's a contradiction.

PS: The same argument works for "if two triangles have two pairs of pairwise equal lengths and also the angle between them is the same then the remaining lengths are equal too". This seems even more reasonable formalization of "T can be moved around while maintaining congruence".