My solution is as follows:
Let $d_1= (a,b) > 0$. Then by Bezout's identity, there exists integers $x_1$and $y_1$ such that, $d_1=ax_1+by_1$. Similarly let $d_2 = (a+c,b) > 0$ then there exists integers $x_2$ and $y_2$ such that, $d_2=(a+c)x_2+by_2$. Then since $b\mid c$ there exists integer $k$ such that $c=bk$ for some $k$.
Note that, $d_2=(a+c)x_2+by_2$. Since $c = bk$ then, $d_2=(a+bk)x_2+by_2$ $\Rightarrow$ $d_2 = ax_2+b(kx_2+y_2)$, which is a linear combination in $a$ and $b$. Since $d_1 = (a,b)$, it divides any linear combination in $a$ and $b$. Thus $d_1\mid d_2$.
Similarly, $d_1=ax_1+by_1$ $\Rightarrow$ $d_1=ax_1+by_1+cx_1-cx_1$. Since, $c=bk$ $\Rightarrow$ $d_1=(a+c)x_1+b(y_1-kx_1)$. Which is a linear combination in $a+c$ and $b$. Thus $d_2\mid d_1$.
Since $d_1$ and $d_2$ are $\gcd$ of two integers, and $d_1\mid d_2$ and $d_2\mid d_1$, we conclude that $d_1=d_2$ as desired.
Thank you for reading upto here. If there is any errors feel free to comment.
Seems fine.
Notice that we have
$$\gcd(a+b,b)=\gcd(a, b)$$
WLOG, if $c=kb$, $k \ge 1, k \in \mathbb{Z}$, then by induction, we have
\begin{align}\gcd(a+c,b)&=\gcd(a+kb, b)\\&= \gcd(a+kb-b, b)\\&=\gcd(a+(k-1)b, b)\\ &= \gcd(a,b) \end{align}