Hint $\rm\qquad a,b\mid m\iff ab\mid am,bm \iff ab\mid \overbrace{(am,bm)}^{\large \color{#c00}{ (a,\,b)\,m}}\iff ab/(a,b)\mid m$
Remark $\ $ If above we employ Bezout's Identity to replace the gcd $\rm\:(a,b)\:$ by $\rm\:j\,a + k\,b\:$ (its linear representation) then we obtain the proof by Bezout in lhf's answer (but using divisibility language). Note the key role played by the gcd distributive law, i.e. $\rm\:\color{#c00}{(a,b)\,c} = (ac,bc).$
This proof is more general than the Bezout proof since there are rings with gcds not of linear (Bezout) form, e.g. $\rm \Bbb Z[x,y]$ the ring of polynomials in $\,\rm x,y\,$ with integer coefficients, where $\,\rm gcd(x,y) = 1\,$ but $\rm\, x\, f + y\, g\neq 1\,$ (else evaluating at $\rm\,x,y = 0\,$ yields $\,0 = 1).\,$
The proof shows that $\rm\ a,b\mid m\iff ab/(a,b)\mid m,\ $ i.e. $\ \rm lcm(a,b) = ab/(a,b)\ $ using the universal definition of lcm. $ $ The OP is the special case $\rm\,(a,b)= 1.$