Coproducts in $\mathsf{Grp}$ The limits and colimits in the category of abelian groups are as nice as can be, since products and equalizers are the same as in the category of sets.
In the category of groups, however, the coproduct is an unpleasant a less manageable creature. One can show that that's just the way things are, but why is this really? Why does the lack of commutative "spoil" things, and what exactly does it spoil?
 A: At least in algebraic contexts, I like to think of the coproduct of $(A,\cdot)$ and $(B,*)$ as being the structure which $A\cup B$ generates* (applying the original multiplication between pairs of adjacent elements of $A$ or of $B$). That is, every element of the coproduct must be writable as a product of $A$'s and $B$'s - like (assuming associativity):
$$a_1b_1a_2b_2\ldots a_nb_n.$$
This is, of course, basically what a free product looks like in the category of groups - and we can't proceed further because the group laws alone do not allow us to move around elements, beyond removing identity elements from the product.
When we're in the category of abelian groups, however, we obviously have that
$$a_1b_1a_2b_2\ldots a_nb_n=a_1a_2\ldots a_n b_1b_2\ldots b_n$$
which means that every element may be written as a product of an element of $A$ and an element of $B$ - which is basically what happens in a direct product - meaning those two structures will coincide. So, we can regard that abelian groups are special in that their additional structure lets us simplify the usually complex form of the coproduct.
(*Of course, $(A,\cdot)$ is not actually the signature of a group - but it's still a working illustration)
A: As you point out, if the sets $A$ and $B$ are abelian groups, then the product and coproduct are the same. This means that the object $A\times B\in AGrp$  is a $\textit coproduct$. and we have the canonical insertions $$i_{1}:A\to A\times B;a\mapsto (a,1)$$ and $$i_{2}:B\to A\times B;b\mapsto (1,b)$$ that satisfy the UMP of the coproduct. 
We will show by counterexample that the same kind of construction does not work in $Grp$:
Take $A=\left \{ 1,a \right \};a^{2}=1$ and $B=\left \{ 1,b \right \};b^{2}=1$ i.e: $A$ and $B$ are copies of the two-element group. If $A\times B$ is a coproduct, and $i_{1}$ and $i_{2}$ are as above, then define $C$ to be the free group on two generators. i.e. $C$ has elements of the form aaabbaabbbbbb, identity $e=$ the empty word and the multiplication is cancatenation. Note that $C$ is not abelian. i.e. $ab\neq ba$.
Now, let $f:A\to C$ be given by $a\mapsto a$ i.e the word $a$ of length $1$ in $C$ and 
$g:B\to C$ be given by $b\mapsto b$. 
If $A\times B$ is to satify the UMP of the coproduct, we should be able to find a group homomorphism $h:A\times B\to C$ such that $$h\circ i_{1}=f$$ and $$h\circ i_{2}=g$$ 
But we may observe that in $A\times B$, we must have $(a,1)(1,b)=(a,b)=(1,b)(a,1)$ so that if such an $h$ existed, we would require
$ab=f(a)g(b)=((h\circ i_{1})(a))((h\circ i_{2})(b))=h(a,1)h(1,b)=h(a,b)=h(1,b)h(a,1)=((h\circ i_{2})(b))((h\circ i_{1})(a))=g(b)f(a)=ba$. 
The foregoing implies that if such an $h$ exists, then $ab=ba$. But $C$ is not abelian, so no such $h$ exists $\Rightarrow A\times B$ is not a coproduct. 
