Below is a proof that works in any GCD domain, using the universal definitions of GCD, LCM. These ideas go back to Euclid, who defined the greatest common measure of line segments. Nowadays this can be viewed more generally in terms of fractional ideals or Krull's $v$-ideals.
Theorem $\rm\ \ \ \left(\dfrac{a}b,\,\dfrac{A}B\right)\: =\: \dfrac{(a,A)}{[b,B]}\ \ $ if $\rm\ \ (a,b) = 1 = (A,B),\ \ $ where $\rm\ \ \begin{eqnarray} (a,b) &:=&\rm\ gcd\rm(a,b)\\\ \rm [a,b]\, &:=&\rm\ lcm(a,b)\end{eqnarray}$
Proof
$\rm\begin{eqnarray} &\rm\quad c &|&\rm a/b,\,A/B \\
\quad\iff&\rm Bbc &|&\rm\ aB,\,Ab \\
\iff&\rm Bbc &|&\rm (aB,\,\color{#C00}A\color{#0A0}b) \\
\iff&\rm Bbc &|&\rm (aB,\, (\color{#C00}A,aB)\,(\color{#0A0}b,aB))\ \ &\rm by\quad (x,\color{#C00}y\color{#0A0}z) = (x,\,(\color{#C00}y,x)\,(\color{#0A0}z,x)),\ \ see\ [1] \\
\iff&\rm Bbc &|&\rm (aB,\, (A,a)\, (b,B))\ \ &\rm by\quad (a,b) = 1 = (A,B) \\
\iff&\rm Bbc &|&\rm (a,A)\,(b,B)\ \ &\rm by\quad (A,a)\ |\ a,\ (b,B)\ |\ B \\
\iff&\rm\quad c &|&\rm (a,A)/[b,B]\ \ &\rm by\quad (b,B)\:[b,B] = bB, \ \ see\ [2]
\end{eqnarray}$
Here are said links to proofs of the gcd laws used: law [1] and law [2].
Note $\,a^2,b^2\mid m\Rightarrow(ab)^2\!\mid m^2\Rightarrow \color{#c00}{ab}\mid m,\,$ by $\,r\in\Bbb Q,\,r^2\in\Bbb Z\Rightarrow r\in \Bbb Z,\,$ by Rational Root Test
Thus $\,\color{#0a0}{a^2,b^2}\mid m\!\iff a^2,\color{#c00}{ab},b^2\mid m^{\phantom{|^|}}\!$
so $\,\ {\rm lcm}(\color{#0a0}{a^2,b^2}) = {\rm lcm}(a^2,ab,b^2)^{\phantom{|^|}}\! $ via LCM Universal Property (or by: above $(\!\!\iff\!\!)$ shows both sides have same set $S$ of common multiples $m,\,$ so the same least common multiple $= \min S$).
Alternatively with $[x,y]:={\rm lcm}(x,y)\,$ & applying LCM distributive law to expand products
$ [a,b][aa,bb] = [aaa,aab,abb,bbb] = [a,b]^3\,$ so $\,[aa,bb] = [a,b]^2\,$ by cancelling $[a,b]\neq 0,\,$
same as in the GCD Freshman's Dream $\,(a^n,b^n) = (a,b)^n = (a^n, a^{n-1}b,\ldots, a b^{n-1}a, b^n)$
Best Answer
If $a=0$, then $\gcd(a,b)=\operatorname{lcm}(a,b)=\lvert b\rvert$ and $\gcd(a,c)=\operatorname{lcm}(a,c)=\lvert c\rvert$, so we get $2\lvert b\rvert = 2\lvert c\rvert \;\;\to\;\; \lvert b\rvert = \lvert c\rvert$. For $a\neq 0$, rearranging your equation results in
$$\operatorname{lcm}(a,b) - \operatorname{lcm}(a,c) = \gcd(a,c) - \gcd(a,b) \tag{1}\label{eq1A}$$
Since $a \mid \operatorname{lcm}(a,b)$ and $a \mid \operatorname{lcm}(a,c)$, this means $a$ divides the LHS above. However, we have $0 \lt \gcd(a,c) \le \lvert a\rvert$ and $0 \lt \gcd(a,b) \le \lvert a\rvert$, so the RHS is $\gt -\lvert a\rvert$ and $\lt \lvert a\rvert$, so since $a$ divides it, this means it must be $0$. Thus, we get
$$\operatorname{lcm}(a,b) = \operatorname{lcm}(a,c), \;\; \gcd(a,c) = \gcd(a,b) \tag{2}\label{eq2A}$$
Using the relationship you mentioned of $\gcd(a,b)\cdot \operatorname{lcm}(a,b) = \lvert ab\rvert$, and \eqref{eq2A} above, gives that
$$\gcd(a,b)\cdot \operatorname{lcm}(a,b) = \lvert ab\rvert = \gcd(a,c)\cdot \operatorname{lcm}(a,c) = \lvert ac\rvert \;\to\; \lvert ab\rvert = \lvert ac\rvert \;\to\; \lvert b\rvert = \lvert c\rvert \tag{3}\label{eq3A}$$