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.
$\rm\begin{eqnarray} \rm(a,\:b)\ (a,\:c) &=&\rm (a(a,\!\:c),b(a,\!\:c)) &=&\rm ((aa,ac),\:(ba,bc)) &=&\rm (aa,ac,\:\!ba,\:bc) \\ \rm\ (a\dot+ b)(a\dot+c) &=&\rm a(a\dot+c)\dot+b(a\dot+c) &=&\rm (aa\dot+ac)\dot+(ba\dot+bc) &=&\rm aa\dot+ac\dot+ba\dot+bc \end{eqnarray}$