Note that any metric with unique infinite geodesics on $\mathbb R^2$ has this property. In particular, hyperbolic plane as noted by Heather Macbeth. It also includes Minkowski plane with smooth ball and all complete negatively curved Riemannian metrics on $\mathbb R^2$.
So it is better to ask:
Is it true that the function $p$ is the only possible? ($p(a,b,c)=0,1,$ or $2$ when the triangle inequality for a,b,c, correspondingly, fails, turns into equality, or is strict.)
Is it true that space has to be homeomorphic to $\mathbb R^2$?
Sketch for locally compact case You condition implies existence of line (i.e. infinite minimizing geodesic) through any pair of points. Such line must be unique; otherwise for some values $p(a,b,a+b)=\infty$ for some values $a,b$.
If space is locally compact then line depends continuousely on the points. For some points $a_1, a_2, a_3, a_4$ take geodesic $a_1a_2$ connect each point to $a_3$ and connect each point of this trianle to $a_4$. This way we get a cone-map of 3-simplex in your space. If it is nondegenerate (i.e. the image does not coinsides with image of its boundary) then it would be easy to get a triple of small numbers $a,b,c$ such that $p(a,b,c)=\infty$ thus we arrive to a contradiction.
Thus the space must be 2-dimensional and with unique infinite geodesics --- I bet it should be homeomorphic to $\mathbb R^2$. Then it follows that $p(a,b,c)=0,1$ or $2$ as defined above.