TITLE: Localizing an arbitrary additive category
QUESTION [11 upvotes]: Under which conditions localizing an additive category by some class S of morphisms yields and additive category? It seems easy to define certain addition on morphisms if we fix their representatives as zig-zags (i.e. compositions of 'old' morphisms with inverses of morphisms in S; here I use the fact that 'my' S is closed with respect to direct sums of morphisms), but I am not sure at all that this addition will not depend on the choice on representatives. Is there any reasonable condition that will ensure this? I definitely do not want to restrict myself to abelian or triangulated categories.
It seems that in the situations I am interested in, any morphism is a composition of the embedding of a direct summand, an inverse of a morphism from S, and an 'old' morphism (i.e. it is 'almost a fraction'). The Ore conditions are not fulfilled (in general, probably); yet some weakening of them could hold.
I would be deeply grateful for any associations here!
My examples are:
For an additive (pseudo-abelian) category B consider some full triangulated (thick) subcategory D of $K^b(B)$; then my S for B is the set of morphisms in B that yield objects of D (if considered as complexes of length 1). 
In particular, S is always closed with respect to compositions and direct sums of morphisms. 
In fact, I am interested in all aspects of this setup!

REPLY [3 votes]: Here is the statement of Exercise 8.4 p. 202 of 
[KS] Categories and Sheaves by Kashiwara and Schapira:

(a) Let $\mathcal C$ be an additive category and $\mathcal S$ a right multiplicative system. Prove that the localization $\mathcal C_{\mathcal S}$ is an additive category and $Q:\mathcal C\to\mathcal C_{\mathcal S}$ is an additive functor. 

I'll describe what I think is the argument Kashiwara and Schapira have in mind. 
Statement (a) follows from 

(b) there is a pre-additive category on $\mathcal C_{\mathcal S}$ making $Q$ additive, 
(c) $Q$ is essentially surjective. 

Statement (b) is easy, and is treated in a very detailed way at the beginning of the following text of Dragan Miličić: 
http://www.math.utah.edu/~milicic/Eprints/dercat.pdf
As (b) is obvious, it only remains to explain why (b) and (c) imply (a). The key is Lemma 8.2.3 (ii) p. 169 of [KS], which says:

(d) Let $\mathcal C$ be a pre-additive category; let $X,X_1,$ and $X_2$ be objects of $\mathcal C$; and, for $a=1,2$, let $X_a\xrightarrow{i_a}X\xrightarrow{p_a}X_a$ be morphisms satisfying 
  $$
p_a\circ i_b=\delta_{ab}\ \operatorname{id}_{X_a},\quad i_1\circ p_1+i_2\circ p_2=\operatorname{id}_X.
$$
  Then $X$ is a product of $X_1$ and $X_2$ by $p_1,p_2$ and a coproduct of $X_1$ and $X_2$ by $i_1,i_2$. 

Proof. For any $Y$ in $\mathcal C$ we have 
$$
\operatorname{Hom}_\mathcal C(Y,p_a)\circ\operatorname{Hom}_\mathcal C(Y,i_b)=\delta_{ab}\ \operatorname{id}_{\operatorname{Hom}_\mathcal C(Y,X_a)},
$$ 
$$
\operatorname{Hom}_{\mathcal C}(Y,i_1)\circ\operatorname{Hom}_{\mathcal C}(Y,p_1)+\operatorname{Hom}_{\mathcal C}(Y,i_2)\circ\operatorname{Hom}_{\mathcal C}(Y,p_2)=\operatorname{id}_{\operatorname{Hom}_{\mathcal C}(Y,X)}.
$$ 
This implies that $\operatorname{Hom}_{\mathcal C}(Y,X)$ is a product of $\operatorname{Hom}_{\mathcal C}(Y,X_1)$ and $\operatorname{Hom}_{\mathcal C}(Y,X_2)$ by $\operatorname{Hom}_{\mathcal C}(Y,p_1),\operatorname{Hom}_{\mathcal C}(Y,p_2)$, and thus, $Y$ being arbitrary, that $X$ is a product of $X_1$ and $X_2$ by $p_1,p_2$, and we conclude by applying this observation to the opposite category. q.e.d. 
(d) implies:

(e) Let $F:\mathcal C\to\mathcal C'$ be an additive functor of pre-additive categories; let $X,X_1,$ and $X_2$ be objects of $\mathcal C$; and, for $a=1,2$, let $X_a\xrightarrow{i_a}X\xrightarrow{p_a}X_a$ be morphisms such that $X$ is a product of $X_1$ and $X_2$ by $p_1,p_2$ and a coproduct of $X_1$ and $X_2$ by $i_1,i_2$. Then $F(X)$ is a product of $F(X_1)$ and $F(X_2)$ by $F(p_1),F(p_2)$ and a coproduct of $F(X_1)$ and $F(X_2)$ by $F(i_1),F(i_2)$. 

In view of (e), (b) and (c) imply (a).