This is Exercise 1.2.2 of "Combinatorial Group Theory: Presentations of Groups in Terms of Generators and Relations," by W. Magnus et al.
The Question:
Show that $$G=\langle a, b, c\mid P, Q, R, \dots , ab=ba, ac=ca, bc=cb\rangle$$ is an Abelian group.
My Attempt:
This is probably a case of something being so glaringly obvious it's hard to know where to start.
The generators commute, so for $g=W_g(a, b, c), h=W_h(a, b, c)\in G$ as words in $a, b, c$, we have $$\begin{align} gh&=W_g(a, b, c)W_h(a, b, c) \\ &=W_h(a, b, c)W_g(a, b, c)\quad(\text{since we can move } W_h\text{ through } W_g) \\ &=hg. \end{align}$$
Is that a valid proof?
This could work more generally, too, right? It's an instance of Abelianisation of $H=\langle \mathcal G\mid \mathcal R\rangle$ by adding the relations $\mathcal A=\{\alpha\beta=\beta\alpha\mid \alpha, \beta\in \mathcal G\}$ to get the Abelian group $$\operatorname{Ab}(H)=\langle \mathcal G\mid \mathcal R\cup \mathcal A\rangle.$$ If the generators commute, the group is always Abelian.
More generally your group $G$ is a quotient of an abelian group ($ \mathbb{Z}^3$) by $P,Q,R, \dots$. Thus there is a surjective homomorphism $\varphi : \mathbb{Z}^3 \to G $ which implies $G$ is abelian : $$ab = \varphi(A)\varphi(B)=\varphi(AB)= \varphi(BA) = \varphi(B)\varphi(A)=ba$$