The opposite of a preadditive category is preadditive

By [L1], for objects $A,B$ the hom-set ${\mathcal{C}}^{\mathrm{o}\mathrm{p}}(A,B)=\mathcal{C}(B,A)$ already carries the abelian group structure given by [L2].

Let $f,g:B\to A$ in $\mathcal{C}$ and let $h:C\to B$, $k:A\to D$. In ${\mathcal{C}}^{\mathrm{o}\mathrm{p}}$ the corresponding arrows are $f^{\mathrm{o}\mathrm{p}},g^{\mathrm{o}\mathrm{p}}:A\to B$, $h^{\mathrm{o}\mathrm{p}}:B\to C$, and $k^{\mathrm{o}\mathrm{p}}:D\to A$. The inherited addition satisfies $f^{\mathrm{o}\mathrm{p}}+g^{\mathrm{o}\mathrm{p}}=(f+g{)}^{\mathrm{o}\mathrm{p}}$. Reversing composition and applying bilinearity gives $h^{\mathrm{o}\mathrm{p}}\circ (f^{\mathrm{o}\mathrm{p}}+g^{\mathrm{o}\mathrm{p}})=((f+g)\circ h{)}^{\mathrm{o}\mathrm{p}}=(f\circ h+g\circ h{)}^{\mathrm{o}\mathrm{p}}=h^{\mathrm{o}\mathrm{p}}\circ f^{\mathrm{o}\mathrm{p}}+h^{\mathrm{o}\mathrm{p}}\circ g^{\mathrm{o}\mathrm{p}}$. Every arrow in this equality goes from $A$ to $C$ in the opposite category. Similarly, $(f^{\mathrm{o}\mathrm{p}}+g^{\mathrm{o}\mathrm{p}})\circ k^{\mathrm{o}\mathrm{p}}=(k\circ (f+g){)}^{\mathrm{o}\mathrm{p}}=(k\circ f+k\circ g{)}^{\mathrm{o}\mathrm{p}}=f^{\mathrm{o}\mathrm{p}}\circ k^{\mathrm{o}\mathrm{p}}+g^{\mathrm{o}\mathrm{p}}\circ k^{\mathrm{o}\mathrm{p}}$, with every arrow going from $D$ to $B$. These are the two required distributive laws.

Thus ${\mathcal{C}}^{\mathrm{o}\mathrm{p}}$ has abelian-group hom-sets and bilinear composition, so it is preadditive.