Localization of an additive category which is no longer additive 
Is there a nice example of an additive category $C$ and a family of morphisms $S\subset Mor(C)$ such that $C[S^{-1}]$ is no longer additive?

I know that in general localization of categories behaves badly (see examples here, Wayback Machine), but I haven't seen an example of $C[S^{-1}]$ not being even additive.
Also, it is known that if $S$ satisfies Ore conditions, then the localization $C[S^{-1}]$ will be additive, and the localization functor $Q\colon C\to C[S^{-1}]$ will also be additive. So one should search for required examples among ones where $S$ is not Ore.
Thank you very much for your help!

Let's let $C$ be the category of complex vector spaces, and $S$ be the set consisting entirely of the projection operator $p:\Bbb C \to 0$.
I claim that the localization $C[S^{-1}]$ is a new category where we've simply imposed an equivalence relation on Hom-sets:
$$
Hom_{C[S^{-1}]}(V,W) = Hom_C(V,W) / \{\text{rank 1 maps} \sim 0\}
$$
First, since a composite of any rank $\leq 1$ map with any other map has rank $\leq 1$, this does define a category $D$.
First, the map in $S$ maps to an isomorphism in $D$.  In this category $D$ the identity and zero map $\Bbb C \to \Bbb C$ are the same, so the projection $p$ is an isomorphism with inverse $0: 0 \to \Bbb C$.
Moreover, any rank $1$ map in $C$ has a factorization $V \stackrel{f}{\to} \Bbb C \stackrel{g}{\to} W$, and in $C[S^{-1}]$ we have:
$$
gf = gp^{-1} p f = g p^{-1} 0 = g p^{-1} p 0 = g 0 = 0.
$$
This actually tells us that any functor $C \to C'$ sending $p$ to an isomorphism actually factors uniquely through $D$.  This universal property makes $D$ into a localization of $C$ with respect to $S$.
However, the addition on $C$ doesn't descend to $D$ at all (any map is a sum of rank-1 maps).
(In this case, we've chosen a localization that doesn't really interact with the additive structure of $C$ at all, which is the source of the problem.  I don't really know if it's in the spirit of the question you asked.)
