I'll be using additive notation for groups regardless of whether the group in question is Abelian or not.
I am thinking about the consequences of adding propertyless square roots of existing elements to groups.
Right now I'm interested in cases where the Abelianness of the group in question is preserved or not preserved.
Let $G$ be the free group generated by $a$.
If we add a new element $b$ such that $b+b=a$, then the group that we get back is isomorphic to $G$.
If our new element $b$ satisfies $b+b = 0$, then it is less clear whether the resulting group is Abelian or not.
I have a proof that I think works using an infinite permutation of the integers.
Let $a$ correspond to the two-place relation $xAy$ defined to be true precisely when $x = y + 1$.
Let $b$ correspond to the two-place relation $xBy$ defined as $B = \{(1, 0), (0, 1)\} \cup \{ (x, y) : x = y \land x \not \in \{0, 1\} \}$, i.e. $b$ swaps $0$ and $1$ and leaves the rest of the elements alone.
I claim that the group generated by $A$ and $B$ is not Abelian.
$2 (A \circ B) 0$ holds but $2 (B \circ A) 0$ fails.
This permutation representation is a valid interpretation of $\langle a, b \mid b^2 = 0 \rangle$, and it serves to show that the freest interpretation of $\langle a, b \mid b^2 = 0 \rangle$ is not Abelian regardless of whether it is the freest interpretation or not. (I think the permutation representation does correspond to the freest interpretation of $\langle a, b \mid b^2 = 0 \rangle$, but I don't know how to prove it.)
Anyway, finding this infinite permutation representation was slightly tricky, so I'm wondering whether there are other ways to show that adjoining an additional square root of $0$ results in a non-Abelian group, especially ones that are more general or do not involve giving a concrete interpretation to the group.
Adding a "propertyless square root" to a given group is equivalent to taking a free product of the given group with the group $\mathbb Z / 2 \mathbb Z$. So, your question is equivalent to asking whether the free product $\mathbb Z * \mathbb Z / 2 \mathbb Z$ is abelian.
The general fact is that any free product of two nontrivial groups is nonabelian. The best proof I know uses Bass-Serre theory to concretely realize the group by an action on a tree ("best" meaning, here, that you learn a very nice and very powerful theory with lots of applications).
If you want a method which does not involve giving a "concrete interpretation" to the group, you're not going to be satisfied with that. But, frankly, representing groups "concretely" is the textbook way to identify many groups, including free products. The general method is to learn something about a mysterious group $G$ by using a presentation for $G$ by generators and relators to construct a "representation" of $G$, meaning a homomorphism $G \mapsto \Gamma$ to some other more well-known group $\Gamma$ (which can be a permutation group, or a matrix group, or anything, really). One reason why this is so useful is that once one correctly guesses how to define the representation on each generator of $G$, each relator can be easily verified to hold in $\Gamma$ by using whatever well known method of computation there might be in the group $\Gamma$.
In this case you can also see more simply that $\mathbb Z * \mathbb Z / 2 \mathbb Z$ is nonabelian by representing it in the group of rigid motions of the real number line, i.e. all functions of the form $x \mapsto \pm x + t$: $$a \to (x \mapsto x+1) \qquad b \ \to (x \mapsto -x) $$ The only relator that needs to be checked is $b^2 = \text{Id}$ (I cannot bring myself to write $b^2=0$... eeerghhh, I wrote it...), and obviously the composition of $x \mapsto -x$ with itself is the identity.
On a side note, this representation of $\mathbb Z * \mathbb Z / 2 \mathbb Z$ is not injective: it satisfies an additional relator $bab=a^{-1}$ which is not satisfied in $\mathbb Z * \mathbb Z/2\mathbb Z$; its image is the infinite dihedral group. If one is interested, there is a more clever injective representation of $\mathbb Z * \mathbb Z / 2 \mathbb Z$ in the group $PSL(2,\mathbb R)$ of fractional linear transformation $x \mapsto \frac{px + q}{rx + t}$ ($pt-qr=1$).