I'll use more convenient notations: in the free abelian group $\mathbf Z^{(X)}$ generated by a set $X$, I'll denote $[x]$ the element $e_x$, i.e. the map which sends $x$ to $1$ and any $x'\ne x$ to $0$.
This being said, you have a bijective map from $\mathbf Z^{(A\times B)}$ to $\mathbf Z^{(B\times A)}$, which sends $[(a,b)]$ to $[(b,a)]$. This map sends generators of the relations defining the tensor product in the first free group, namely $[(a_1,b)]+[(a_2,b)]-[(a_1+a_2,b)]$ onto the generators of the relations defining the tensor product in the second free group, $[(b,a_1)]+[(b,a_2)]-[(b,a_1+a_2)]$, hence the subgroup $R_{A\times B}$ generated by the first set onto the subgroup $R_{B\times A}$ generated by the second group.
Therefore we have a commutative diagram of abelian groups
\begin{alignat}{5}
0\longrightarrow &R_{A\times B}\hookrightarrow&&\mathbf Z^{A\times B}\longrightarrow A\otimes B\longrightarrow 0 \\
&\quad\downarrow&&\enspace\downarrow\\
0\longrightarrow &R_{B\times A}\hookrightarrow&&\mathbf Z^{A\times B}\longrightarrow B\otimes A\longrightarrow 0
\end{alignat}
which induces a morphism from $A\otimes B$ to $B\otimes A$ by the universal property of kernels. As the the two vertical maps are group isomorphisms, the induced morphism is an isomorphism too.
The most elegant way to solve such problems with no effort is to use universal properties and in particular the Yoneda Lemma. In fact, this Lemma tells us that two objects with the same universal properties are isomorphic. A universal property of an object $R$ is just a description of the homomorphisms $R \to S$ for any other object $S$ of the ambient category (or, of the homomorphisms $S \to R$). Here, we are in the category of commutative $\mathbb{Q}$-algebras (of course, you could also work in the category of rings, but it would make life less easy).
So if $S$ is a commutative $\mathbb{Q}$-algebra, then the universal properties of quotient rings ("fundamental theorem on homomorphisms") and polynomial algebras ("evaluation homomorphisms") show us that there are natural bijections
$\mathrm{Hom}(\mathbb{Q}[r,s]/(r+s-1,\, r-r^2,\, s-s^2),S)\\
\cong \{(a,b) \in S^2 : a+b-1=0,\, a-a^2=0,\, b-b^2=0\},$
$\mathrm{Hom}(\mathbb{Q}[y]/(y^2-1),S) \cong \{c \in S : c^2-1=0\}.$
So all we have to do is to find a natural bijection between the sets
$\{(a,b) \in S^2 : a+b=1,\, a=a^2,\, b=b^2\} \cong \{c \in S : c^2=1\}.$
But notice that $a+b=1$ means that $b$ is superfluous, we may replace it by $1-a$. The equation $b=b^2$ then holds automatic, since in general if $a$ is idempotent ($a=a^2$) then $1-a$ is idempotent. (This also shows, by the way, that the ideal $(r+s-1,r-r^2,s-s^2)$ is equal to the ideal $(r+s-1,r-r^2)$.)
So all we have to show is that the set of idempotents in $S$ is naturally isomorphic to the set of involutions in $S$ (i.e. $c^2=1$).
Well, if $a$ is idempotent, then $c := 1-2a$ is an involution, since $c^2=1-4a+4a^2=1$. Conversely, if $c$ is an involution, then $a := (1-c)/2$ is idempotent, since $a^2=(1-2c+c^2)/4=(2-2c)/4=(1-c)/2=a$. These maps are inverse to each other. This finishes the proof.
This correpondence between idempotents and involutions can also be motivated or explained with a bit of algebraic geometry. An idempotent is nothing but a section on $\mathrm{Spec}(S)$ with values in $0,1$ (in the residue fields). An involution is nothing but a section on $\mathrm{Spec}(S)$ with values in $-1,+1$. Now we just need a linear transformation which takes $\{0,1\}$ to $\{-1,+1\}$, for example $1-2x$, to get the correspondence.
Here is the whole proof in one chain of natural isomorphisms:
$\mathrm{Hom}(\mathbb{Q}[r,s]/(r+s-1,r-r^2,s-s^2),S)\\
\cong \{(a,b) \in S^2 : a+b-1=0,\, a-a^2=0,\, b-b^2=0\}, \qquad | ~ \text{ substitute } b=1-a \\
\cong \{a \in S : a=a^2,\, (1-a)=(1-a)^2=0\} \\
= \{a \in S : a=a^2\} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad ~\,\, | ~ \text{ substitute } a=(1-c)/2 \\
\cong \{c \in S : (1-c)/2=\bigl((1-c)/2)\bigr)^2\} \\
= \{c \in S : c^2=1\} \\
\cong \mathrm{Hom}(\mathbb{Q}[y]/(y^2-1),S).$
Best Answer
Let $\varphi: \mathbb{Z}[X] \to \mathbb{Z}\left[\sqrt{-3}\right]$ be the unique ring homomorphism sending $X$ to $\sqrt{-3}$. It is not difficult to show that $\varphi$ is surjective and its kernel is the ideal $(X^2+3)$. By the first isomorphism theorem, $\varphi$ induces a ring isomorphism $$\overline{\varphi}: \frac{\mathbb{Z}[X]}{(X^2+3)} \longrightarrow \mathbb{Z}\left[\sqrt{-3}\right]$$ sending $X+(X^2+3)$ to $\sqrt{-3}$.
Now consider the ideal $$\frac{(2+X,X^2+3)}{(X^2+3)}=\left(2+X+(X^2+3)\right)$$ of $\frac{\mathbb{Z}[X]}{(X^2+3)}$. What does it corresponds to under $\overline{\varphi}$? We compute it: $$\overline{\varphi}\left(\frac{(2+X,X^2+3)}{(X^2+3)}\right) = \overline{\varphi}\left(2+X+(X^2+3)\right)=\left(2+\sqrt{-3}\right).$$ We conclude that $$\frac{\frac{\mathbb{Z}[X]}{(X^2+3)}}{\frac{(2+X,X^2+3)}{(X^2+3)}} \cong \frac{\overline{\varphi}\left(\frac{\mathbb{Z}[X]}{(X^2+3)}\right)}{\overline{\varphi}\left(\frac{(2+X,X^2+3)}{(X^2+3)}\right)} = \frac{\mathbb{Z}\left[\sqrt{-3}\right]}{\left(2+\sqrt{-3}\right)}.$$