Answer
Yes.
Work Step by Step
The vectors are clearly linearly independent. (because there are $2$ and they are not parallel)
Say $(a,b)=k(1,1)+l(-1,1)$, then $a=k-l$ and $b=k+l$, thus $k=0.5(a+b)$ and $l=0.5(b-a)$, thus every $2$ dimensional vector is expressable, thus it is a basis.
