Let's use the universal definition of $A\otimes B$: $(A\otimes B,\iota)$, where $\iota\colon A\times B\to A\otimes B$ is a bilinear (resp. $R$-bilinear) map, is the unique group (resp. $R$-module) with the property that for any bilinear map $\phi\colon A\times B\to C$ (resp. $R$-bilinear), where $C$ is any abelian group (rep. $R$-module), there exists a unique homomorphism $\Phi\colon A\otimes B\to C$ such that $\phi=\Phi\circ\iota$.
Let $s\colon A\times B\to B\times A$ be the map $s(a,b) = (b,a)$, and let $(B\otimes A,j)$ be the tensor product of $B$ and $A$. Note that $B\otimes A$ has the corresponding universal property relative to $B\times A$ and $j$.
Now, the map $js\colon A\times B\to B\otimes A$ is a bilinear map (composition of a homomorphism and a bilinear map). Therefore, there exists a unique $\mathcal{F}\colon A\otimes B\to B\otimes A$ such that $js = \mathcal{F}\iota$. Likewise, the map $\iota s^{-1}\colon B\times A\to A\otimes B$ is bilinear, so there exists a unique $\mathcal{G}\colon B\otimes A\to A\otimes B$ such that $\iota s^{-1} = \mathcal{G}j$.
Now consider $\mathcal{GF}\colon A\otimes B\to A\otimes B$. We have that $\mathcal{GF}\iota = \mathcal{G}js = \iota s^{-1}s = \iota.$ But there is supposed to be a unique map $f\colon A\otimes B\to A\otimes B$ such that $f\circ\iota = \iota$ (since $\iota$ is bilinear), and clearly $f=\mathrm{id}_{A\otimes B}$ works. Therefore, $\mathcal{GF}=\mathrm{id}_{A\otimes B}$.
Symmetrically, $\mathcal{FG}j = \mathcal{F}\iota s^{-1} = jss^{-1} = j$. But $j$ is a bilinear map $B\times A\to B\otimes A$, so there is supposed to be a unique map $g\colon B\otimes A\to B\otimes A$ such that $gj=j$. Since $\mathrm{id}_{B\otimes A}$ works, that is the unique function with the desired property. Since $\mathcal{FG}$ also works, $\mathcal{FG}=\mathrm{id}_{B\otimes A}$.
Thus, $\mathcal{FG}=\mathrm{id}_{B\otimes A}$ and $\mathcal{GF}=\mathrm{id}_{A\otimes B}$. Therefore, $\mathcal{F}\colon A\otimes B\to B\otimes A$ is an isomorphism, as desired.
Note that we don't need to know how we represent $A\otimes B$; we just need the universal property (and that a tensor product exists for any [ordered] pair of groups). Though one can likewise use the universal property to show that if $(M,\iota)$ is a tensor product for $A\times B$, then $(M,\iota s^{-1})$ is a tensor product for $B\times A$, so you just need to know $A\otimes B$ exists.