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.