Let $k$ be a field, $I$ and $J$ infinite sets, and $A$ the $k$-subalgebra of $$k(t)[x_i,y_i: i\in I,j\in J]$$ 
generated by 
$$\{x_i,y_i,tx_i,t^{-1}y_i: i\in I, j\in J\}.$$

Then 
$$(tx_i)_{i\in I}\otimes(t^{-1}y_j)_{j\in J}-(x_i)_{i\in I}\otimes(y_j)_{j\in J}$$ 
is a non-zero element of the kernel of the natural map $A^I\otimes_A A^J\to A^{I\times J}$. 

As your proof shows, this means that $A^I\otimes_AA^I$ must have torsion, and indeed, for any $k\in I$,
$$\begin{align}x_k\left((tx_i)_{i\in I}\otimes(t^{-1}y_j)_{j\in J}\right)
&=(tx_kx_i)_{i\in I}\otimes(t^{-1}y_j)_{j\in J}\\
&=(x_i)_{i\in I}\otimes(x_ky_j)_{j\in J}\\
&=x_k\left((x_i)_{i\in I}\otimes(y_j)_{j\in J}\right),\end{align}$$
and so
$$x_k\left((tx_i)_{i\in I}\otimes(t^{-1}y_j)_{j\in J}-(x_i)_{i\in I}\otimes(y_j)_{j\in J}\right)=0.$$