A model category which is an additive category Let $\cal C$ be a model category which is also additive. Suppose that the homotopy category $\operatorname{Ho}\mathcal C$ is additive, for example this is true when the weak equivalences in $\cal C$ is closed under biproducts (see this question).
If we take a cofibrant object $X$ and a fibrant object $Y$ then there is a natural isomorphism
$$
 \operatorname{Ho}\mathcal C(X,Y) \cong \mathcal C(X,Y)/\sim
$$
where $\sim$ is the homotopy relation. Is this always a group isomorphism?
 A: Yes, because the projection functor from the model category to the homotopy category preserves coproducts of cofibrant objects. That is actually the way of showing that the homotopy category has coproducts.
A: If by "additive" you mean an $\mathbf{Ab}$-enriched category with a zero object and biproducts, then yes. Let $\mathcal{M}$ be model category that is additive in this sense, let $\mathcal{M}_c$ be the full subcategory of cofibrant objects, let $\mathcal{M}_f$ be the full subcategory of fibrant objects, and let $\mathcal{M}_{cf} = \mathcal{M}_c \cap \mathcal{M}_f$. Here are the relevant facts:
$\DeclareMathOperator{\Ho}{Ho}$


*

*The coproduct of a family of cofibrant objects is automatically a homotopy coproduct, so the localising functor $\mathcal{M}_c \to \Ho \mathcal{M}$ preserves coproducts. Dually, the localising functor $\mathcal{M}_f \to \Ho \mathcal{M}$ preserves products.

*Hence, the localising functor $\mathcal{M}_{cf} \to \Ho \mathcal{M}$ preserves the zero object and biproducts. (Note that $\mathcal{M}_{cf}$ is an additive subcategory of $\mathcal{M}$.)

*A category with a zero object and biproducts is automatically enriched over commutative monoids in a unique way, and a functor that preserves zero objects and biproducts is similarly enriched. Thus, there is a unique enrichment of $\Ho \mathcal{M}$ over commutative monoids that makes the localising functor $\mathcal{M}_{cf} \to \Ho \mathcal{M}$ an enriched functor.

*Since $\mathcal{M}_{cf}$ is actually $\mathbf{Ab}$-enriched and the localising functor $\mathcal{M}_{cf} \to \Ho \mathcal{M}$ is full and essentially surjective on objects, $\Ho \mathcal{M}$ is also $\mathbf{Ab}$-enriched. 


Now, let $X$ and $Y$ be any two objects in $\mathcal{M}$. In order for the hom-set map
$$\mathcal{M}(X, Y) \to \Ho \mathcal{M}(X, Y)$$
to be a group homomorphism, it is sufficient that the localising functor $\mathcal{M} \to \Ho \mathcal{M}$ preserve either the coproduct $X + X$ or the product $Y \times Y$. (We already know that it preserves initial and terminal objects.) Thus it suffices to take either $X$ cofibrant or $Y$ fibrant.
