Does $G\oplus G \cong H\oplus H$ imply $G\cong H$ in general? In this question, The Chaz asks whether $G\times G\cong H\times H$ implies that $G\cong H$, where $G$ and $H$ are finite abelian groups. The answer is to his question is yes, by the structure theorem for finite abelian groups, as noted in the answer by Anjan Gupta.
Even though I don't know the first thing about categories -- except for the things that I do know -- I'm wondering if and how such property could be expressed and proven in terms of universal properties, without actually manoevring inside the objects. For instance one may attempt to create a morphism $G\to H$ somehow appealing to the universal property of $\oplus$, and subsequently show this morphism is an isomorphism by chasing diagrams. But it seems likely that the existence of a structure theorem of some sort will be required.
This question may be considered trivial or weird for someone who's fluent with categories, I don't know. This topic contains quite a few references. I haven't really worked through any of them (yet) but I couldn't find anything helpful at first sight.
-- edit, So a more precise title would have been "Under what conditions that can be expressed in a universal way does $G\oplus G\cong H\oplus H$ imply $G\cong H$ ?" but I don't like long titles.
-- edit2, By the comment by Alexei Averchenko, maybe It's more natural to ask this question with the product instead of the coproduct. An answer to my question with the 'real' product would be appreciated too, obviously.
 A: Even in the category of all abelian groups, $G\oplus G\cong H\oplus H$ does not imply $G\cong H$: there are countably generated torsion-free abelian groups $G$ such that $G\not\cong G\oplus G$, but $G\cong G\oplus G\oplus G$. Setting $H=G\oplus G$ gives an example where the implication does not hold. (The first examples were constructed by A.L.S. Corner, On a conjecture of Pierce concerning direct decompositions of abelian groups, Proc. Colloq. Abelian Groups (Tihany, 1963), Akadémiai Kiadó, Budapest, 1964, pp. 43-48; he proves that for any positive integer $r$ there exists a countable torsion-free abelian group $G$ such that the direct sum of $m$ copies of $G$ is isomorphic to the direct sum of $n$ copies of $G$ if and only if $m\equiv n\pmod{r}$.)
Of course, the analogous result for direct products fails for groups, though it holds for groups having both chain conditions (by the Krull-Schmidt theorem). 
The fact that you have categories with arbitrary coproducts/products (like the category of all groups and the category of all abelian groups) in which the result fails means that no proof via universal properties alone can exist. A proof that depended only the universal properties would translate into any category in which products/coproducts exist, but the result is false in general even for very nice categories. 
Whether the conditions for the implication to hold can be expressed via universal conditions also seems to me to be doubtful. Universal conditions don't lend themselves easily to statements of the form "If something happens for $A$ and $B$, then it happens for $f(A)$ and $f(B)$" (where by "$f(-)$" I just mean "this other object cooked up or related to $A$ in some way") unless the construction/relation happens to be categorical.
