Observe that $\bigoplus_I : \textbf{Ab}^I \to \textbf{Ab}$ is a conservative exact functor: it is right exact by general nonsense, it preserves monomorphisms (because e.g. $\bigoplus_{i \in I} A_i$ is naturally a subgroup of $\prod_{i \in I} A_i$), and it is conservative because $\bigoplus_{i \in I} A_i \cong 0$ implies each $A_i \cong 0$.
It seems to me that you are looking for a conservative exact functor $\mathcal{A} \to \textbf{Ab}$, so this observation implies that allowing $I$ with more than one element does not add any generality.
For small abelian categories $\mathcal{A}$, the Freyd–Mitchell embedding theorem gives a fully faithful exact functor $\mathcal{A} \to R\textbf{-Mod}$ (where $R$ is a not necessarily commutative ring), so every small abelian category admits a conservative exact functor $\mathcal{A} \to \textbf{Ab}$.
In particular, $\mathcal{A}$ will be isomorphic to a (not necessarily full) abelian subcategory of $\textbf{Ab}$, by which I mean a subcategory that is closed under finite direct sums/products, kernels, and cokernels.
If $\mathcal{A}$ is not required to be small then one has to do more work.
The existence of a conservative exact functor $\mathcal{A} \to \textbf{Ab}$ is itself a size restriction on $\mathcal{A}$.
For example:
Proposition.
If there is a conservative exact functor $\mathcal{A} \to \textbf{Ab}$, then $\mathcal{A}$ is locally small and wellpowered.
Proof.
A conservative exact functor is automatically faithful, and $\textbf{Ab}$ is locally small, so $\mathcal{A}$ must also be locally small.
Similarly, a conservative exact functor induces embeddings of subobject lattices, and $\textbf{Ab}$ is wellpowered, so $\mathcal{A}$ must also be wellpowered. ◼
Maybe you don't believe in categories that are not locally small and wellpowered.
Even so, I am not aware of any embedding theorems that work for arbitrary locally small and wellpowered abelian categories.
(I am also not aware of counterexamples.
Perhaps there is some large cardinal axiom that implies it can be done.)
In practice, it is not necessary to embed the entire category to prove the theorems you want – you can usually find a small abelian subcategory containing all the objects and morphisms you need for your theorem and then you can embed that subcategory.
Asking for an embedding of the whole category at once is being greedy – like asking for a global holomorphic chart of a complex manifold when local charts suffice.
Here are some embedding theorems I know for non-small abelian categories.
Theorem.
If $\mathcal{A}$ is a Grothendieck abelian category then there is a conservative exact functor $\mathcal{A}^\textrm{op} \to \textbf{Ab}$ that has a left adjoint.
Proof.
It is a well-known theorem that Grothendieck abelian categories have injective cogenerators.
But an injective cogenerator of $\mathcal{A}$ is precisely an object $I$ such that $\textrm{Hom}_\mathcal{A} (-, I) : \mathcal{A}^\textrm{op} \to \textbf{Ab}$ is a conservative exact functor.
Furthermore, representable functors in cocomplete categories automatically have a left adjoint, and $\mathcal{A}^\textrm{op}$ is indeed cocomplete. ◼
Maybe contravariance is jarring.
But $\textbf{Ab}$ is itself a Grothendieck abelian category, so the theorem (or Pontryagin duality) gives us a conservative exact functor $\textbf{Ab}^\textrm{op} \to \textbf{Ab}$, and composing them yields a conservative exact functor $\mathcal{A} \to \textbf{Ab}$.
The following is a small generalisation.
Theorem.
If $\mathcal{A}$ is a locally small abelian category and has a small generating set, then there is a conservative exact functor $\mathcal{A}^\textrm{op} \to \textbf{Ab}$.
Proof.
The hypothesis implies there is a small full abelian subcategory $\mathcal{B}$ containing the given small generating set of $\mathcal{A}$.
We get a fully faithful functor $\mathcal{A} \to [\mathcal{B}^\textrm{op}, \textbf{Ab}]$, but in any case it is not automatically exact.
Let $\textbf{Lex} (\mathcal{B}, \textbf{Ab})$ be the full subcategory of left exact functors $\mathcal{B}^\textrm{op} \to \textbf{Ab}$.
Then, the earlier functor factors as a fully faithful exact functor $\mathcal{A} \to \textbf{Lex} (\mathcal{B}, \textbf{Ab})$.
But $\textbf{Lex} (\mathcal{B}, \textbf{Ab})$ is a Grothendieck abelian category, so we may apply the earlier theorem to conclude. ◼
