I have a question: Let $n\in\mathbb{Z}$, then $(a+bn,b)=(a,b)$, $(a,b)$ being the greatest common divisor.
Is this a correct approach?
Let $e=(a+bn,b), \quad n\in\mathbb{Z}$, then $e$ is expressible as the smallest linear combination of $an+b$ and $b$ so there exists $u,v\in\mathbb{Z}$ such that $$ e=u(a+bn)+vb $$ and $e$ is the smallest positive such number. Rearranging on the right hand side we get $$ e=ua+(un+v)b $$ which gives a linear combination of $a$ and $b$. Since $e$ is still the smallest such we get that $(a,b)=e$. If this is correct is there a more elegant solution?
That doesn't sound quite convincing -- when you say "since $e$ is still the smallest such" it sounds to me like you're essentially assuming what you were supposed to prove.
I suspect your strategy is to show that the integer combinations of $a+bn$ and $b$ are the same as the integer combinations of $a$ and $b$. But in order to do that without handwaving you need to show that every combination of $a+bn$ and $b$ is also a combination of $a$ and $b$, and that every combination of $a$ and $b$ is also a combination of $a+bn$ and $b$. Your proof as it stands only does the first half of this.
(An alternative which seems more direct to me would be to show that the common divisors of $a+bn$ and $b$ are the same as the common divisors of $a$ and $b$).