Your arguments work just fine, but we can do a bit better.

Let's define $$G=\langle a,b\mid a^2=b^3=a^{-1}b^{-1}ab=e\rangle.$$

As you noted, $a^{-1}b^{-1}ab=e$ if and only if $ab=ba.$ Hence, $a$ and $b$ commute, and so since $G$ is generated by $a$ and $b,$ then it follows immediately that $G$ is an abelian group, and in particular $$G=\{a^mb^n\mid m,n\in\Bbb Z\}.$$ Since $a^2=e$ and $b^3=e,$ then we can conclude that $$G=\{e,a,b,ab,b^2,ab^2\}.$$

At this point, there are a few ways we can proceed:


Show that $e,a,b,ab,b^2,ab^2$ are distinct, then note that the abelian group of order $6$ is cyclic.
Show that $e,a,b,ab,b^2,ab^2$ are distinct, then apply Structure Theorem for Finite Abelian groups, as you did, together with the fact that $\gcd(2,3)=1,$ so that $G$ is cyclic.
Show that $e,a,b,ab,b^2,ab^2$ are distinct, then demonstrate an element of $G$ of order $6.$
Show that $H:=\langle a\rangle\cong\Bbb Z_2,$ $K:=\langle b\rangle\cong\Bbb Z_3,$ and that $H\cap K=\{e\},$ so since $H,K$ are normal subgroups of $G$ (since $G$ abelian) and $HK:=\{hk\mid h\in H,k\in K\}=G,$ then $G\cong H\times K.$ (Have you seen this result before?) Thus, $G\cong\Bbb Z_2\times\Bbb Z_3\cong\Bbb Z_6.$


That last is arguably the simplest. It suffices to show that $a\ne e,$ $b\ne e,$ $b^2\ne e,$ $a\ne b,$ and $a\ne b^2.$

Cyclic group of order $6$ defined by generators $a^2=b^3=a^{-1}b^{-1}ab=e$.
I have a Compositing node set-up for a lomography effect that I want to export or append to other blend files. Is there a way to do this? 

I tried grouping the nodes, but the group wasn't appearing anywhere when I tried to append them in. (Node setup attached for whatever it's worth.)
