It is not true in general that if $\gcd(x,y)=d$ then $(x)+(y)=(d)$. In particular, if $R=\mathbb{Z}[x]$ (the ring of polynomials over $\mathbb{Z}$ in the indeterminate $x$), then $\gcd(2,x)=1$ (since any divisor of $2$ in $R$ must be an integer, and the only integers that divide $x$ in $R$ are $\pm 1$), however $(2,x)=(2)+(x)\subsetneq R$.
More generally, integral domains for which gcds can be written as a linear combination are known as Bézout domains--domains where every finitely generated ideal is principal.
As Bill mentions in his answer, GCD domains (ie integral domains in which every pair of nonzero elements has a gcd) satisfy the property that every irreducible is prime. In particular, if $R$ is an integral domain in which every nonzero nonunit can factor into irreducibles, then $R$ is a GCD domain if and only if $R$ is a UFD.
As for showing that $6$ and $2+2\sqrt{-5}$ have no gcd, this can also be done easily enough using the norm function $N:\mathbb{Z}[\sqrt{-5}]\rightarrow\mathbb{N}\cup\{0\}$ defined by $N(a+b\sqrt{-5})=a^2+5b^2$. In particular for all $\alpha,\beta\in R$, $N(\alpha\beta)=N(\alpha)N(\beta)$, $N(\alpha)=0$ if and only if $\alpha=0$, and $N(\alpha)=1$ if and only if $\alpha\in U(R)$ (proving this is a canonical abstract algebra exercise). So, if there is a gcd of 6 and $2+2\sqrt{-5}$, then there must be a common divisor of 3 and $1+\sqrt{-5}$ (since 2 divides both 6 and $2+2\sqrt{-5}$ in $R$).
Thus, there exists $\alpha\in R$ such that $\alpha \mid 3$ and $\alpha \mid 1+\sqrt{-5}$. Taking the norm (and using the norm properties above), we have $N(\alpha)\mid 9$ and $N(\alpha) \mid 6$. Therefore, it follows that either $N(\alpha)=1$ or $N(\alpha)=3$. However, the fact that there are no integer solutions to the equation $a^2+5b^2=3$ (why?) implies that there is no element in $R$ having a norm of 3. Therefore $N(\alpha)=1$ and $\gcd(3,1+\sqrt{-5})=1$.
Now suppose that $6$ and $2+2\sqrt{-5}$ have a gcd. Then we have $2=2\gcd(3,1+\sqrt{-5})=\gcd(6,2+2\sqrt{-5})$. However, we reach a contradiction as $1+\sqrt{-5}$ is a common divisor of both 6 and $2+2\sqrt{-5}$, but $1+\sqrt{-5}$ does not divide 2 (Hint: suppose it does and use the norm). Therefore $\gcd(6,2+2\sqrt{-5})$ does not exist.
We give a proof of the type you were writing, but with the logical details made more explicit.
Theorem: (Bézout's Identity) Two integers $m$ and $n$ are relatively prime if and only if there exist integers $s$ and $t$ such that $ms+nt=1$.
Because $a$ and $b$ are relatively prime, there exist integers $x$ and $y$ such that $ax+by=1$.
Since $d\mid a$, there is an integer $k$ such that $a=dk$.
It follows that $d(kx)+by=1$. Thus (taking $s=kx$ and $t=y$) we conclude that $d$ and $b$ are relatively prime.
The second problem is more awkward if we want to use linear combinations, but it can be done that way. Since $a$ and $b$ are reatively prime, we have $ax+by=1$ for some integers $x$ and $y$.
Now consider linear combinations $as+ct$ of $a$ and $c$. We have $acx+bcy=c$, and therefore
$$as+ct=as+(acx+bcy)t=a(s+cxt)+bc(yt).$$
Thus any linear combination of $a$ and $c$ is a linear combination of $a$ and $bc$.
The converse is obvious: any linear combination of $a$ and $bc$ is a linear combination of $a$ and $c$.
It follows that the set of linear combinations of $a$ and $c$ is the same as the set of linear combinations of $a$ and $bc$. Thus in each case the smallest positive linear combinations are the same, and hence the gcd's are the same.
Best Answer
Let $a$ and $b$ be coprime. Then $[a]_b$ generates $\mathbb Z/b\mathbb Z$. So there is some $x$ such that $x[a]_b=[1]_b$. By definition, this means there exists a $y$ such that $xa-1=yb$, as desired.