Biproducts are associative, commutative, and unital up to canonical isomorphism

By [L1], both $A\oplus (B\oplus C)$ and $(A\oplus B)\oplus C$ are products of the ordered triple $(A,B,C)$. Let $\alpha :A\oplus (B\oplus C)\to (A\oplus B)\oplus C$ be the unique morphism whose three composites with the target projections are the three source projections, and define $\beta$ in the reverse direction the same way. Then $\beta \alpha$ and $1_{A\oplus (B\oplus C)}$ have the same three composites with the source projections, while $\alpha \beta$ and $1_{(A\oplus B)\oplus C}$ have the same three composites with the target projections. By the product universal property in [L3], $\alpha$ and $\beta$ are inverse isomorphisms. This is the canonical associativity isomorphism.

Likewise, $A\oplus B$ and $B\oplus A$ are products of the same ordered pair after swapping the labels. The unique maps exchanging the two projections are inverse by the same product-uniqueness argument, giving the canonical symmetry isomorphism $A\oplus B\cong B\oplus A$.

By [L2], the object $0$ is both initial and terminal. Let $p_A:A\oplus 0\to A$ and $p_0:A\oplus 0\to 0$ be the product projections from [L1], and let $\lambda :A\to A\oplus 0$ be the unique map with $p_A\lambda =1_A$ and $p_0\lambda$ equal to the unique map $A\to 0$. Then $p_A$ and $\lambda$ are inverse, because $p_A\lambda =1_A$ and $\lambda p_A$ has the same composites with $p_A$ and $p_0$ as $1_{A\oplus 0}$. So $A\oplus 0\cong A$. The same argument with the product projections of $0\oplus A$ gives $0\oplus A\cong A$.

The three displayed isomorphisms are therefore forced by the universal properties alone, which is exactly the asserted associativity, commutativity, and unitality up to canonical isomorphism.