Isomorphism of direct sums $R/a \oplus R/b \cong R/{\rm lcm}(a,b)\oplus R/\gcd(a,b)$

Hint $ $ The Bezout identity $\rm\: ad+bc\:\! =\:\! g\:$ yields the following Smith normal form reduction $\left[\begin{array}{cc}\rm a & \!\!\!0\\0 &\rm \!\!\!b\end{array}\right] \!\sim\! \left[\begin{array}{cc} \rm a &\rm \!\!\!b\\0 &\rm\!\!\! b\end{array}\right] \!\sim\! \left[\begin{array}{cc} \rm a &\rm\!\!\! b\\0 &\rm\!\!\! b\end{array}\right] \left[\begin{array}{cc} \rm d &\rm \!\!\!\smash{-\!b/g} \\ \rm c &\rm a/g\end{array}\right] $ $= \left[\begin{array}{cc} \rm g &\rm 0\\ \rm bc &\rm \!\!\!ab/g\end{array}\right]^{\phantom{|^|}} \!\sim\! \left[\begin{array}{cc} \rm g &\rm 0\\ 0 &\rm \!\!ab/g\end{array}\right] = \left[\begin{array}{cc} \rm \!gcd(a,b) &\rm 0\\0 &\rm \!\!\!\!\!\!\!\smash{lcm(a,b)}\!\end{array}\right] $

See this answer for further detail.

I'm quite confident that your proof is optimal. In any case I think it's a pretty proof - you should be satisfied!