Addition of roofs makes an additive localization

For $(s,f),(t,g):X\to Y$, choose $a\in S,b$ with $sa=tb=r\in S$ and define their sum as $(r,fa+gb)$. If roofs already have denominator $r$, they agree exactly when their numerators agree after some further refinement $e$ with $re\in S$: one implication is immediate; for the other, cancel $r$ between the two refinement legs and then precompose by the resulting denominator.

For two choices of common denominator, refine those denominators once more. Equality of each of the two summands can then be witnessed at that denominator by the previous criterion; applying cancellation successively makes both numerator equalities simultaneous. Their sums are equal there by additivity. This proves choice and representative independence. At a common denominator, associativity, commutativity, zero and negation are precisely the abelian-group laws in $\mathcal{C}(U,Y)$.

Postcomposition by an ordinary arrow is additive since it acts on numerators. Precomposition by an ordinary arrow is additive: use one Ore square with the common denominator of both summands. Composition with $Q(s{)}^{-1}$ is the inverse of the additive composition bijection for $Q(s)$, and is therefore additive. Every localized arrow is a product of ordinary arrows and inverse denominators, proving bilinearity.

The image of $1_0=0$ shows $Q(0)$ is a zero object: every arrow to or from it is zero by bilinearity and the identity law. For $B=X\oplus Y$, the equations $p_X{i}_X=1$, $p_Y{i}_Y=1$, $p_X{i}_Y=p_Y{i}_X=0$ and $i_X{p}_X+i_Y{p}_Y=1_B$ survive under the additive $Q$. They supply unique tuples of incoming and outgoing arrows, hence make $Q(B)$ a biproduct. The empty biproduct is $Q(0)$, and binary ones iterate to finite ones.