Exactness of $A\to A\amalg B\underset{(0,1)}\leftrightarrows B$ In a pointed category, the sequence $A\overset{(1,0)}\to A\times B\overset{(0,1)}\leftrightarrows B$ is always exact.
However, the "dual" sequence $A\to A\amalg B\underset{(0,1)}\leftrightarrows B$ is generally far from exact. Its exactness hints at a commutative situation. Consider for instance the category of monoids. The arrow $(0,1)$ acts on a reduced word via 
$bab^\prime\mapsto bb^\prime$. If $bb^\prime=1\in B$ then $bab^\prime\in \operatorname{Ker}(0,1)\setminus A$, so the sequence is not exact.
Does the exactness of $A\to A\amalg B\underset{(0,1)}\leftrightarrows B$ in fact imply $A\amalg B\cong A\times B$? If not, what does it imply along this direction?
Update. As Eric Wofsey's answer points out, this is not true e.g for pointed sets. What if the category is assumed unital?
 A: In the category of pointed sets, $A\to A\amalg B\to B$ is always exact, but coproducts and products are not isomorphic.  What you can say is that if $A\to A\amalg B\to B$ is always exact, then the natural map $A\amalg B\to A\times B$ always has trivial kernel.  Indeed, the kernel of $A\amalg B\to A\times B$ is just the pullback of the kernels of the two component maps $A\amalg B\to A$ and $A\amalg B\to B$, which by assumption are the inclusions $i:A\to A\amalg B$ and $j:B\to A\amalg B$.  Now suppose $f:C\to A$ and $g:C\to B$ are maps such that $if=jg$.  Composing $if=jg$ with the map $(1,0):A\amalg B\to A$ we find $f=(1,0)if=(1,0)jg=0$.  Similarly, $g=0$.  This shows that the pullback of $i$ and $j$ is $0$, as desired.
As for a unital example, I believe you can take the category of all monoids with the property that $xy=1$ implies $x=1$ and $y=1$.  Such monoids are closed under limits and coproducts, and $A\to A\amalg B\to B$ is always exact for them.  The map $A\amalg B\to A\times B$ is not typically an isomorphism for such monoids though: it is surjective and has trivial kernel, but it is not injective.
