Let $a,k,m$ be integers. Prove that $\gcd(ka,km) = k\gcd(a,m)$.

You seem to be on the right track, although I'm not so sure about your last step. Also, I would avoid using fractions when possible. Here's my approach.

Let $g = \gcd(a,m)$ and let $d = \gcd(ka,km)$. We want to show that $d = kg$.

Since $d = \gcd(ka,km)$, we know by the Extended Euclidean Algorithm that $d = (ka)s + (km)t$ for some $s,t \in \Bbb Z$. But then observe that $d = k(as + mt)$. Hence, it suffices to prove that $g = as + mt$.

Since $g = \gcd(a,m)$, we know by the Extended Euclidean Algorithm that $g$ is the smallest possible integer that can be expressed in the form $ax + my$, where $x,y \in \Bbb Z$. But since $as + mt$ can also be expressed in this form, we know that $\boxed{g \leq as + mt}$.

Since $g=ax+my$, we may scale this equation by $k$ to obtain $kg = (ka)x + (km)y$. But recall that $d$ is the smallest possible integer that can be expressed in the form $(ka)s + (km)t$, where $s,t \in \Bbb Z$. Hence, we have $k(as + mt)=d \leq kg$ so that cancelling the $k$ from both sides yields the inequality $\boxed{as + mt \leq g}$ (we assume, without loss of generality, that $k\geq0$).

Finally, since $g \leq as + mt$ and $as + mt \leq g$, we have shown that $g=as+mt$, as desired.

This is proven using Bezout's Identity in this answer.

Bezout says that $$ \gcd(a,b)=\inf\{ax+by\gt0:x,y\in\mathbb{Z}\} $$ Then we simply note that $$ \inf\{cax+cby\gt0:x,y\in\mathbb{Z}\} $$ is both $c\gcd(a,b)$ and $\gcd(ac,bc)$.

I believe this is very similar to your proof, but I think you need to exploit the fact that $\gcd(a,b)$ is the smallest positive $ax+by$ for any $x,y\in\mathbb{Z}$.