Let $k$ be a field, $I$ and $J$ infinite sets, and $A$ the $k$-subalgebra of $$k(t)[x_i,y_j: i\in I,j\in J]$$ generated by $$\{x_i,y_j,tx_i,t^{-1}y_j: i\in I, j\in J\}.$$
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 $A^I\otimes_A A^J\to A^{I\times J}$.
[EDIT: Here's a proof that this element is non-zero.
If it's zero, then it can be shown to be zero using only finitely many elements of $A$, and so $(tx_i)_{i\in I}\otimes(t^{-1}y_j)_{j\in J}-(x_i)_{i\in I}\otimes(y_j)_{j\in J}=0$ in $A^I\otimes_BA^J$ for some finitely generated subalgebra $B$ of $A$. We can choose $r\in I$ and $s\in J$ so that $$B\subseteq k(t)[x_i,y_j:i\neq r, j\neq s].$$
Now $A$ has a basis consisting of elements of the form $t^lm$ where $m$ is a monomial in $\{x_i,y_j:i\in I,j\in J\}$ and $l\geq 0$ if $m$ involves no $y_j$s and $l\leq0$ if $m$ involves no $x_i$s. The basis elements other than those for which $m$ is a power of $x_r$ or of $y_s$ span an ideal. Let $\bar{A}$ be the corresponding quotient algebra; then the image $\bar{B}$ of $B$ in $\bar{A}$ is just $k$.
Consider the image of $(tx_i)_{i\in I}\otimes(t^{-1}y_j)_{j\in J}-(x_i)_{i\in I}\otimes(y_j)_{j\in J}$ in $\bar{A}^I\otimes_{\bar{B}}\bar{A}^J$. This is non-zero since $tx_r\otimes t^{-1}y_s-x_ry_s\neq0$ in $\bar{A}\otimes_{\bar{B}}\bar{A}$, as the set $\{x_r,tx_r,y_s,t^{-1}y_s\}$ is linearly independent in $\bar{A}$, and the components for $i\neq r$ or $j\neq s$ are all zero.]
As your proof shows, this means that $A^I\otimes_AA^J$ must have torsion, and indeed, for any $s\in I$, $$\begin{align}x_s\left((tx_i)_{i\in I}\otimes(t^{-1}y_j)_{j\in J}\right) &=(tx_sx_i)_{i\in I}\otimes(t^{-1}y_j)_{j\in J}\\ &=(x_i)_{i\in I}\otimes(x_sy_j)_{j\in J}\\ &=x_s\left((x_i)_{i\in I}\otimes(y_j)_{j\in J}\right),\end{align}$$ and so $$x_s\left((tx_i)_{i\in I}\otimes(t^{-1}y_j)_{j\in J}-(x_i)_{i\in I}\otimes(y_j)_{j\in J}\right)=0.$$
