The commutative-monoid enrichment of a category with finite biproducts is unique

Fix $f,g:A\to B$, and write ${\mathrm{\nabla }}_B:B\oplus B\to B$ for the codiagonal and $i_1,i_2:B\to B\oplus B$ for the coproduct injections. Because composition is bilinear for $⊞$, the morphism $i_{1}f⊞i_{2}g:A\to B\oplus B$ satisfies $p_1(i_{1}f⊞i_{2}g)=f$ and $p_2(i_{1}f⊞i_{2}g)=g$, where $p_1,p_2$ are the product projections and the off-diagonal terms vanish by the biproduct zero equations. Thus $i_{1}f⊞i_{2}g=⟨f,g⟩$ by the product universal property.

Applying ${\mathrm{\nabla }}_B$ and using bilinearity again gives ${\mathrm{\nabla }}_B⟨f,g⟩={\mathrm{\nabla }}_B(i_{1}f⊞i_{2}g)={\mathrm{\nabla }}_B{i}_{1}f⊞{\mathrm{\nabla }}_B{i}_{2}g=f⊞g$, since ${\mathrm{\nabla }}_B{i}_1={\mathrm{\nabla }}_B{i}_2=1_B$. But the left-hand side is exactly the canonical sum $f+g$ from [L1].

Therefore $f⊞g=f+g$ for every pair of parallel morphisms. So every bilinear commutative-monoid enrichment compatible with the same finite biproducts is forced to equal the canonical one.