Since you say you cannot use the simple min/max exponents proof via unique factorization, here is a proof that uses only universal gcd laws (so will work in any gcd domain). We simply eliminate all lcms by $\rm\:[x,y] = xy/(x,y),\:$ and apply gcd laws (distributive, commutative, associative, etc).
$$\rm\begin{eqnarray}
\rm &\rm\qquad\qquad (a,[b,c])\ &=&\rm\ [(a,b),(a,c)] \\
\rm \iff&\rm\qquad\quad \left(a,\dfrac{bc}{(b,c)}\right)\ & =&\rm\ \dfrac{(a,b)(a,c)}{(a,b,c)} \\
\iff &\rm (a,b,c)(a(b,c),bc)\ &=&\rm\ (a,b)(a,c)(b,c)
\end{eqnarray}$$
which is true since both sides $\rm = (aab,aac,abb,abc,acc,bbc,bcc)\:$ by distributivity etc.
If you are not proficient with gcd laws, you may find it helpful to rewrite the proof employing a more suggestive arithmetical notation, namely denoting the gcd $\rm (a,b)\:$ by $\rm\ a \dot+ b.\:$ Because the arithmetic of GCDs shares many of the same basic laws of the arithmetic of integers, the proof becomes much more intuitive using a notation highlighting this common arithmetical structure. Below is a sample calculation comparing the two notations.
$$\rm\begin{eqnarray}
\rm(a,\:b)\ (a,\:c) &=&\rm (a(a,\!\:c),b(a,\!\:c)) &=&\rm ((aa,ac),\:(ba,bc)) &=&\rm (aa,ac,\:\!ba,\:bc) \\
\rm\ (a\dot+ b)(a\dot+c) &=&\rm \color{#c00}{a(a\dot+c)}\dot+b(a\dot+c) &=&\rm (\color{#c00}{aa\dot+ac})\dot+(ba\dot+bc) &=&\rm aa\dot+ac\dot+ba\dot+bc
\end{eqnarray}$$
Now the gcd arithmetic looks like integer arithmetic, e.g. uses of the gcd distributive law look the same as for integers, e.g. $\ \rm \color{#c00}{a(a\!+\!c) = aa\!+\!ac}\,$ etc.
Hint $\rm\,\ m,n\mid x\!\iff\! mn\mid mx,nx\!\iff\! mn\mid\overbrace{(mx,nx)}^{\textstyle (m,n)x}\!$ $\rm\iff\! \overbrace{mn/(m,n)}^{\textstyle \ell}\mid x$
Let $\rm\:x = \ell\:$ in $\,(\Leftarrow)\,$ to get $\rm\:m,n\mid \ell,\:$ i.e. $\rm\:\ell\:$ is a common multiple of $\rm\:m,n,\:$ necessarily the least common multiple, since $(\Rightarrow)$ shows $\rm\:m,n\mid x\:\Rightarrow\:\ell\mid x\:\Rightarrow\:\ell\le x.$
Remark $\ $ That $\rm\: m,n\mid x\iff \ell\mid x\,$ is a definition of $\rm\,{\rm lcm}(m,n)\,$ in more general rings since - as above - it implies that $\,\ell\,$ is a common multiple of $\rm\,a,b\,$ that is divisibly least, i.e. it divides every common multiple. See this answer for this universal approach to LCMs and GCDs.
Best Answer
Your proof is correct, but you need not use the decomposition of an integer into prime factors. More elementary arguments work here.
As you said, because $\operatorname{gcd}(a,b)=1$, any common multiple of $a$ and $b$ is in fact a multiple of $ab$. Conversely, any multiple of $ab$ is a common multiple of $a$ and $b$.
Hence, the least common multiple must be $ab$ itself.
NB: to justify that any common multiple of $a$ and $b$ is in fact a multiple of $ab$, we can proceed by using Bézout's theorem. We are given a relation $$au+bv=1$$ where $u,v$ are integers. Now, let $m$ be a common multiple of $a$ and $b$. Let us write $m=aa'=bb'$. Then $$m=aa'=(a^2u+abv)a'=(aa')au+(ab)a'v=(bb')au+(ab)a'v=ab(b'u+a'v)$$ With this, the proof is complete with only elementary arguments.
NB2: As @abc... stated, it is in general true that $ab=\operatorname{gcd}(a,b)\operatorname{lcm}(a,b)$. We can now prove this without using the decomposition into prime factors, thanks to the fact that if $k$ is any integer, then $k\operatorname{gcd}(a,b)=\operatorname{gcd}(ka,kb)$ and $k\operatorname{lcm}(a,b)=\operatorname{lcm}(ka,kb)$.
Indeed, given $a,b$, it follows from this property than $\operatorname{gcd}(\frac{a}{\operatorname{gcd}(a,b)},\frac{b}{\operatorname{gcd}(a,b)})=1$. Applying the proven result, we know that $\operatorname{lcm}(\frac{a}{\operatorname{gcd}(a,b)},\frac{b}{\operatorname{gcd}(a,b)})=\frac{ab}{\operatorname{gcd}(a,b)^2}$. Now, multiply both sides by $\operatorname{gcd}(a,b)^2$ to get the result.