Abelian objects in a category 
If $\mathcal{C}$ is a category, $A\in\mathcal{C}$ is an abelian object if $\mathcal{C}(-,A)$ is naturally an abelian group. Assuming $\mathcal{C}$ has enough limits, this is equivalent to there being a multiplication morphism $m:A\times A\to A$, an identity $\epsilon:*\to A$ and an inverse $i:A\to A$, such that all the diagrams commutes.

I take this to mean $C(T,A)$ is an abelian group for every $T\in\text{Ob}(\mathcal{C})$. Is that the correct interpretation? How then does 'having enough limits' induce this?
(Secondary curiosity, if anyone knows, would this mean that the $H$-spaces are the abelian objects in the category of pointed topological spaces, with morphisms being homotopy classes of maps.)
 A: The part "$\mathcal C(-,A)$ is a naturally abelian group" means that there exists a functor $\mathbf A : \mathcal C^{\rm op} \to \mathsf{Ab}$ such that that $U \circ \mathbf A$ and $\mathcal C(-,A)$ are isomorphic (as functors $\mathcal C^{\rm op} \to \mathsf{Set}$) where $U$ is the forgetful functor $\mathsf{Ab} \to \mathsf{Set}$.
The part "$\mathcal C$ has enough limits" is just to justify the existence of the product $A\times A$ and of the terminal object $\ast$ in use just after.

Now that the vocabulary is set, the proof is more or less tautological. Given maps $\epsilon, i, m$ making the wanted diagrams commute, you can defined a "pointwise" abelian group structure on $\mathcal C(X,A)$: the unit is $X\to \ast\overset \epsilon\to A$, the inverse operation is taking $f: X \to A$ to $X\overset f\to A \overset i\to A$, and the multiplication takes $f,g: X\to A$ to $X\overset\delta\to X\times X \overset {f\times g}\to A\times A \overset m \to A$. You need to check that it is indeed an abelian group, and that it is functorial.
Conversely, given that "$\mathcal C(-,A)$ is naturally an abelian group", you get natural transformations $$\boldsymbol\epsilon :\mathcal C(-,\ast) \to \mathcal C(-,A) \\ \mathbf i:\mathcal C(-,A) \to \mathcal C(-,A) \\ \mathbf m : \mathcal C(-,A\times A)\simeq \mathcal C(-,A)\times\mathcal C(-,A) \to \mathcal C(-,A)$$ defined at component $X$ as the unit, the inverse and the multiplication of $\mathbf A(X)$ respectively. You need to check that this is actually natural. By Yoneda lemma, it gives $\epsilon, i, m$ such as wanted. The fact that they make all the wanted diagrams commute will be given by $\mathbf A(X)$ being an abelian group for each $X$.
