Since you asked about the "thought process" in Igor's answer, I will show how we can view it simply as a special case of the well-known fact that lines are uniquely determined by two distinct points on the line. In particular this means that any two lines through two distinct points are identical, so have equal slopes.
By hypothesis all $(\ge 2)$ integer points $\,(x,y)\,$ on $\, a y = -bx + d\,$ also lie on the line $\,y = -x + 1\,$ so these lines are identical, so they have equal slope, i.e. $\,-b/a = -1,\,$ so $\,b = a,\,$ QED.
The proof in Igor's answer amounts to equating the slopes $\,m,m'\,$ of each line computed from their two common points $(x,y)$ and $(x',y') = (x+a,y-b),\,$ yielding
$$ m = \dfrac{x'-x}{y'-y} = m'$$
Remark $ $ We don't need to known the values of the two common integer points $\,(x,y),\,(x',y')\,$ on the lines - only that at least two such points exist. But general linear theory tells us infinitely many integer solutions exist for $\,a y +b x = d,\,$ viz. the general solution is the sum of any particular solution plus the general sum of the associated homogeneous equation $\,ay + bx=0.\,$ A particular solution exists by Bezout's identity for the gcd, and the homogeneous equation has an obvious solution $(x,y) = (a,-b),\,$ so adding it to any particular solution yields a second solution.
If you know linear algebra you may find it instructive to rephrase it in that language, e.g. using the fact that the determinant $\,m'-m\,$ of the system $\, y = m x + b,\ y = m' x + b'\,$ must be zero if the solution is not unique.
Best Answer
Ignoring the equations involving $r$ and $s$, we have everything we need. Observe that: \begin{align*} q &= xa + yb \\ zq &= xaz + (yz)b \\ zq &= xaz + (xk)b \\ zq &= x \cdot \underbrace{(az + kb)}_{\in ~ \mathbb Z} \end{align*}