Assume either $a=0$ or $b=0$. If $a=0$, then $\gcd(a,b)=\gcd(0,b)=b$ and $\gcd(a-b,b)=\gcd(-b,b)=-b$, so $\gcd(a,b)\neq\gcd(a-b,b)$.
If $b=0$, then $\gcd(a,b)=\gcd(a,0)=0$ and $\gcd(-b,b)=\gcd(a,0)=0$.
So, what can I say? Can you help? Can you check my proof-trying?
Let l = gcd(a,b) and r = gcd(a-b,b). Then l divides a and b and r divides a-b and b. l divides all linear combinations of a and b so l divides a-b, thus l divides r. Also r divides all linear combinations of a-b and b, mainly r | (a-b)+b = a and thus r divides l. Since $r | l$ and $l | r$ we have that $l=r$.
Let $\delta=\gcd(a,b)$ thus $\delta|\gcd(b,a-b)$. Also $\gcd(b,a-b)|b+(a-b)=a$ therefore $\gcd(b,a-b)|\delta$ and the claim follows
Alternative solution via Bezout's theorem:
Notice $sa+tb = s(a-b) + (s+t)b$ So the linear combinations of $a-b$ and $b$ are precisely the same as the linear combinations of $a$ and $b$.
