Is $\mathbf{Rel}$ pre-additive? I first fix some notation: 
the objects are given by Ob Rel = Ob Set. The morphisms are given by $\hom(A,B)= \mathcal{P}(A \times B)$. Let $R \colon A \to B$ and $S \colon B \to C$. Composition is given by $(a,c) \in S \circ R \iff \exists b ~ \text{s.t.} ~ (a,b) \in R ~ \text{and} ~ (b,c) \in S$.
This category, correct me if I am wrong, is semi-additive with both product and coproduct given by disjoint union.
I wish to know if it is additive. Can one enrich the hom-sets with a group structure bilinear with respect to composition? I propose the following group structure:
$R+S := (R \cup S) - (R \cap S)$
 A: Given a category $\mathcal{C}$ with a zero object and a biproduct (i.e. coproduct canonically isomorphic to product), there is (up to isomorphism) at most one additive structure on $\mathcal{C}$. Indeed, given $f, g : X \to Y$, we must have:
$$f + g = \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} f & 0 \\ 0 & g \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \nabla \circ (f \oplus g) \circ \Delta$$
In particular, for $\mathbf{Rel}$, this gives $\cup$ as the binary operation on hom-sets. This is not cancellative, so there is no way of making $\mathbf{Rel}$ into an additive category.
A: You can think about morphisms in $\text{Rel}$ as being matrices over a particular semiring, namely the truth semiring $\{ 0, 1 \}$. This is the semiring of subsets of a one-element set, with addition given by union (not symmetric difference!) and multiplication given by intersection. Equivalently, it's the semiring $\text{End}(1)$ in $\text{Rel}$.
This semiring is manifestly not a ring, since its addition operation is not cancellative, so there's no reason to expect additivity. The same reasoning applies to categories of matrices over more general semirings. 
