In a commutative ring, GCD exists implies LCM exists?

Write $a=\alpha d$, $b=\beta d$, $d(u\alpha+v\beta-1)=0$, $\mu=\alpha\beta d$.

Let $c \in (a) \cap (b)$. Then $c=as=bt=dc’$ for $s,t \in A$. Thus $c\beta \in (\mu)$, $c\alpha \in \mu$, thus $uc\alpha+vc\beta=c’d(\alpha u+\beta v)=c’d=c \in (\mu)$. The reverse inclusion ($(\mu) \subset (a) \cap (b)$) is clear.

The trick is to effectively "cancel" $d$ at the start (vs. end) of proof. More precisely, since $\,d\mid a\,$ there exists $\,\frac{a}d\in R\,$ with $\,\frac{a}d\,d = a,\,$ and similarly for $\frac{b}d,\,\frac{n}d\,$ by $\,d\mid a,b\mid n.\,$ Fixing your proof this way:

$$\begin{align} (dn)&=((a)+(b))(n)=\,(an)\,+\,(bn)\subset (ab)\ \ \ \text{upon "cancelling"}\ d\\[.4em] {\rm we\ get}\ \ \color{#c00}{(n)} &= ((\frac{a}d)\!+\!(\frac{b}d))\:\!\color{#0a0}n = (\frac{a}d\,n)\! +\! (\frac{b}d\,n) \subset (\frac{a}d\,\frac{b}d\,d), \ \ {\rm by}\ \ a,b\mid n\\ {\rm and\ since\ \ \ } &\,\overbrace{((\frac{a}d)\!+\!(\frac{b}d))\, \color{#0a0}{d\,\frac{n}d}} = (\underbrace{(a)+(b)}_{\large (d)})\,\frac{n}d = \color{#c00}{(n)}.\ \ \ {\rm QED} \end{align}\qquad$$