Answer: $(a, b) = (0, 1)$, $(1, 2)$, $(2, 2)$ and $(-n, n)$ for any integer $n \ge 0$.
If $a+b=0$ then $(a,b) = (-n,n)$ for an integer $n \ge 0$, so now assume that $a+b \ne 0$. Then $a+b = a^2-ab+b^2$ and hence $(a-b)^2 + (a-1)^2 + (b-1)^2 = 2$. By setting $x = a-1, y = b-1$ we get $x \le y$ and $x^2+y^2+(x-y)^2 = 2$. Clearly, $(x,y) = (\pm1, \pm1)$, $(0,1), (-1,0)$, and therefore $(a,b) = (0,1), (1,2), (2,2)$.