I’m assuming from your diagrams that addition kinda includes subtraction, so yes, this seems to be a necessary and sufficient condition! If gcd(a,b) >1, then clearly any point with a coordinate not divisible by this gcd is impossible to reach.
Also if a and b are both odd, if we start at a point whose coordinates are the same parity then adding any of the vectors will give another coordinate with the same parity, so the condition is necessary to get to those with one odd and one even coordinate.
Now let’s show it’s sufficient:
Given the basis shown and field Z for scalars, where can we get to?
We can make the vectors (0,2a), (0,2b), (2a, 0), (2b, 0) by adding (b,a) and (-b, a) and similar, and by Bezout’s theorem we can get the vectors (0,2) and (2,0) iff a, b are coprime. From here we only need to get from (0,0) to any coordinate with the parities (o,e), (e,o) and (o,o).
If one of a or b is even and the other odd, wlog let the odd one be a. Then (b,a) and (a,b) are the vectors to get us from (0,0) to (e,o) and (o,e) respectively, and (b, a)+(a,b) gets us to (o,o). From there we can get anywhere using our (0,2) and (2,0).