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?
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.