Cartesian Plane Travelling. Given that we move from $P_1:(a,b)$ to $P_2:(x,y)$ while moving to $(s\pm t,t)$ or $(s,t\pm s)$ from $(s,t)$ at any instant. We are trying to find if we can reach $P_2$ from $P_1$.
I say that we are always at a point whose both coordinates are linear combination of $(a,b)$, the initial point, though I'm not sure if the associated transformations are independent. Anyways the final point will be reached, therefor, if both $x$ and $y$ are linear combinations of $a$ and $b$, hence divisible by ${\rm HCF}(a,b)$.
Though I have not provided any rigorous proof, am I correct in concluding this or is this wrong, and if yes then why?
 A: The ability to walk from one point to another is an equivalence relation! (If I am mistaken in this all fails ^^)
(Reflexivity) From each point you can walk to the point itself by taking no steps. 
(Symmetry) If you can walk from A to B then you can walk from B to A. This since (s + t,t) together with (s - t,t) and (s, t + s) together with (s, t - s) and constitute inverse operations.
(Transitivity) If you can walk from A to B and from B to C then you can walk from A to C by concatenating the step instructions. 
This gives us the idea of trying to describe what points you can walk between by constructing equivalence classes with a characteristic representative and it will especially be the idea of symmetry that will dominate our reasoning. 
Let $(a,b)$ be a point. This point is in the same class as $(d,d)$ where $d = \gcd(a,b)$ because you can reach this point by means of an implementation of the  euclidean algorithm.
Example:
(144,21) -> (123, 21) -> (102, 21) -> (81, 21) -> (60, 21) -> (39, 21) ->(18, 21) -> (18,3) -> (15,3)->(12,3)->(9,3) ->(6,3)->(3,3)
If a or b are negative and the algorithm is carried in a way that gives the negative (-d,-d) this doesn't change anything as you'll just have to do some rearrangement operations at the end of the chain(-d, -d) ->(-d,0) ->(-d,d) ->(0,d)->(d,d)
So we now have these classes $[(n,n)]$ and we would be done if we could show that there was no overlap between them. I'm sure you can use some corollary of of euclidean algorithm for that but I will opt for a simple but problem specific idea.
If you start from $(n,n)$ then any point you get to will be of the form $(pn,qn)$ and you can clearly not get to a point $(m,m)$ where $0 < m < n$. 
But say you could get from $(n,n)$ to a point $(pn,pn)$, $p>1$. Well that would be absurd because by the previous paragraph you cannot get from $(pn,pn)$ to any point of the form $(m,m)$ where the components are positive and smaller than $pn$ which $n$ would be. 
Therefore the $[(n,n)]$ $(n>0)$ are disjoint. 
In conclusion you can get from $(a,b)$ to $(x,y)$ if and only if $\gcd(a,b) = \gcd(x,y)$.
