Let $x, y \in \mathbb Z$ not both zero.
(a) First prove for any $a,b \in \mathbb Z$, that $ \gcd(x,y)| ax + by$.
(b) Deduce that for any $a, b, c, d \in \mathbb Z$, since similarly, $\gcd (x,y) | cx + dy$, we must have $\gcd(x,y) | \gcd (ax+by, cx+dy)$.
(c) Finally conclude that if, in addition, $ad-bc = \pm 1$ then $\gcd(x,y) = gcd (ax+by, cx + dy)$. The fact that $\gcd(x,y) = \gcd (x+ky,y)$ that we use for Euclidean Algorithm is a very special case of this exercise.
(Edit : I have proved part (a) and I understand part (b), only need help with (c) now)
You have proved part a,which is good.
On part b, all you need to do is apply part a with new integers.
We know that gcd$(x,y) | ax+by$ and gcd$(x,y) | cx+dy$.
Now, the point is that for any $m$ and $n$, gcd$(x,y) | m(ax+by) + n(cx+dy)$.
From Bezout's theorem, we know that gcd($ax+by,cx+dy$) $= m(ax+by)+ n(cx+dy)$ for some $m,n$.
Since the above applies for all $m,n$, it follows that gcd$(x,y) | $gcd$(ax+by,cx+dy)$.
Edit: Part c is in town.
So suppose that $ad-bc = 1$. Then note that $ x= (ad-bc)x + (bd-bd)y=d(ax+by)+ -b(cx+dy)$, so $x$ can be written as a linear combination of the two elements $ax+by$ and $cx+dy$. Similarly, $y= -c(ax+by) +a(cx+dy)$, so $y$ can be written as a linear combination of $ax+by$ and $cx+dy$ too. Hence, any multiple $mx+ny$ can be written as a linear combination of $ax+by$ and $cx+dy$. Therefore, certainly $gcd(x,y)$ can be written as a linear combination, say:$$\gcd(x,y) = m(ax+by) + n(cx+dy)$$
But then, note that $\gcd(ax+by,cx+dy)$ divides the right side! Hence it must divide the left side, so that $\gcd(ax+by,cx+dy)$ divides $\gcd(x,y)$, and hence since we already know that $\gcd(x,y)$ divides $\gcd(ax+by,cx+dy)$,we have that $\gcd(x,y)=\gcd(ax+by,cx+dy)$.
Your work: use similar reasoning when $ad-bc=-1$to get the answer.