Necessary and sufficient condition for $can : A^X\otimes_A A^Y\rightarrow A^{X\times Y}$ to be an embedding The two sets are, of course, supposed infinite.
This question is related to that one
Commutation of tensor products with inverse limits in a specific case
where it received a (partial) answer ($A$ noetherian is suff).
I can complement a bit: in any case, $A$ is supposed a commutative ring with unit and $1\not= 0$.
When $A$ is a domain (no zero divisor) and if $A^X\otimes_A A^Y$ is torsion-free (which I do not know
in general) then the natural arrow  $f\otimes g\mapsto ((x,y)\mapsto f(x)g(y))$
$$
A^X\otimes_A A^Y\rightarrow A^{X\times Y}
$$
is an embedding.

Let $\Phi$ be the natural arrow $A^X\otimes_A A^Y\rightarrow A^{X\times Y}$ and  $t\in ker(\Phi)$. Among all expressions
$$
\alpha\, t=\sum_{i=1}^n f_i\otimes g_i\ , 
$$
with $\alpha\not=0$ choose one with $n$ minimal.


(H) We suppose that the tensor product $A^X\otimes_A A^Y$ is torsion-free


if $n=0$, one has $\alpha\, t=0$ and then $t=0$, we are done. If $n>0$, the family $(g_i)_{i=1}^n$ is free and for all $(x,y)\in X\times Y$
$$
\Phi(\alpha\, t)[x,y]=\sum_{i=1}^n f_i(x)g_i(y)
$$
which means that $(\forall x\in X)(\sum_{i=1}^n f_i(x)g_i=0)$ and then $(\forall x\in X)(\forall i\in [1..n])(f_i(x)=0)$. This implies $\alpha\, t=0$ and then, under (H), $t=0$.

Who knows more ? (I am specially interested in necessary and sufficient conditions but, of course, any partial further result is welcome).
On the way I cannot prove (H), any kind of hint will be appreciated. Hence

Subquestion We know that, if $A^X\otimes_A A^Y$ is torsionless (as a $A$-module), the ring $A$ is a domain. Is the converse true ? otherwise can one provide a counterexample ?

 A: 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
$$x_s\left((tx_i)_{i\in I}\otimes(t^{-1}y_j)_{j\in J}-(x_i)_{i\in I}\otimes(y_j)_{j\in J}\right)=0.$$
