A more natural problem would be to decide when the canonical morphism $\bigoplus_i A_i \to \prod_i A_i$ is an isomorphism. Clearly, this is the case if almost all $A_i$ are trivial. Notice that often when people say that $X,Y$ are not isomorphic, they really mean that a certain canonical morphism $X \to Y$ is not an isomorphism.
The question if there is any "random" isomorphism between $\bigoplus_i A_i$ and $\prod_i A_i$ is not that natural, because this isomorphism may just ignore the inclusions and the projections of the direct sum resp. direct product and therefore can be "wild".
Let us replace for the moment the category of abelian groups with the category of vector spaces over some fixed field $k$. Let $V_i$ be an infinite-dimensional vector space. Then it is known (MO/49551) that $\dim(\prod_i V_i)=\prod_i |V_i|$, where $|V_i|$ is the cardinality of (the underlying set of) $V_i$. Clearly, we also have $\dim(\oplus_i V_i) = \sum_i \dim(V_i)$. It is also well-known (math.SE/194281) that $|V_i|=\max(\dim(V_i),|k|)$. So the question if $\bigoplus_i V_i$ and $\prod_i V_i$ are isomorphic really only depends on the cardinal number equation $\sum_i d_i = \prod_i \max(d_i,q)$, where $q=|k|$ and $d_i=\dim(V_i)$. (The whole algebraic structure has disappeared!) Let us assume for the moment that $k$ is countable, i.e. $q \leq \aleph_0$. Then the equation becomes $\sum_i d_i = \prod_i d_i$. If $d:=d_i$ is constant and $|I|=\aleph_0$, the equation becomes $d = d^{\aleph_0}$. And this is perfectly possible, for example when $d=2^{\aleph_\alpha}$. Of course there are also examples when $d_i$ is not constant and where $I,k$ are arbitrary. (On the other hand, for some cardinals $d$, the equation $d=d^{\aleph_0}$ might as well be independent from ZFC!)
So you see, there are many families of infinite-dimensional vector spaces $V_i$ with $\bigoplus_i V_i \cong \prod_i V_i$, for example $V_i=\mathbb{R}$ over the base field $\mathbb{Q}$. In particular, their underlying abelian groups are isomorphic, contradicting the claim. (Notice that you cannot really write down this isomorphism, and therefore it is of no practical importance.)
It is simply the subgroup of all elements $(a_i)$ in the product $\displaystyle\prod_{i\in\mathbf Z}A_i$ such that all $a_i$ are $0$ but a finite number.
You may think of the construction of the ring of polynomials with coefficients in $R$; it is the $R^{(\mathbf N)}$ of sequences of coefficients almost all $0$.
Best Answer
Yes, you are right. There is a notion of direct sums of (infinitely many) groups, see e.g. Wikipedia, though in my eyes it could be an ambiguous term, as one usually uses speaks about coproducts when people talk about direct sums.
You are also right that this definition does not satisfy the universal property of coproducts. As you noted, the coproduct in the category of groups $\mathbf{Grp}$ is given by the free product. Since coproducts (of the same components) are canonically isomorphic, surely your notion is not the coproduct in $\mathbf{Grp}$.
That is, the forgetful functor $\mathbf{Ab} \to \mathbf{Grp}$ is not cocontinuous (i.e. does not preserve small colimits) which in particular shows that it does not admit a right adjoint.