= Solution
A <pointed category> has a <zero object>, both initial and terminal. Factoring through that object gives a <zero morphism> $0_{A,B}$ between any two objects. In this paper a <semi-additive structure> means <commutative-monoid enrichment>: every hom-set is a <commutative monoid> with additive zero, and composition distributes over addition in both variables. No existence of finite <products in a category> is included in this last definition. This convention matters for the final one-object example; under the stronger convention requiring finite <biproducts>, that example would not be a <semi-additive category>.
Let $c_{A,B}:A+B\to A\times B$ be the given canonical isomorphism. Transfer the coproduct injections across $c_{A,B}$, so $A\times B$ is also a <coproduct in a category>, with injections
$$
i_1=\langle1_A,0\rangle,\qquad i_2=\langle0,1_B\rangle.
$$
Write $\nabla_B:B\times B\to B$ for the <morphism> whose composites with $i_1,i_2$ are both $1_B$. For $f,g:A\to B$, define the <biproduct-induced addition of morphisms> by
$$
\boxed{f+g=\nabla_B\langle f,g\rangle=[1_B,1_B]c_{B,B}^{-1}\langle f,g\rangle.}
$$
The zero is the existing <zero morphism>. We verify the laws from the universal properties, without assuming addition in advance.
All finite canonical maps from coproducts to products are isomorphisms, by induction from the binary ones and the <zero object>. Thus both $(B\times B)\times B$ and $B\times(B\times B)$ are ternary <coproducts in a category>. The two iterated fold maps to $B$ agree on each of the three injections, hence are equal. Applying this to $\langle f,g,h\rangle$ gives associativity. The interchange $B\times B\to B\times B$ swaps the two injections, so its composite with $\nabla_B$ is $\nabla_B$, giving commutativity. Finally $\langle f,0\rangle=i_1f$, so $f+0=f$, and similarly $0+f=f$.
Precomposition is additive because $\langle f,g\rangle u=\langle fu,gu\rangle$. For $v:B\to C$, the <morphisms> $v\nabla_B$ and $\nabla_C(v\times v)$ agree on both injections, so they agree; consequently $v(f+g)=vf+vg$. Composition with a <zero morphism> is zero. This proves the <commutative-monoid enrichment>.
It is unique. In any such enrichment, the additive zero morphisms agree with the pointed ones, since each map to or from the <zero object> belongs to a singleton hom-set. The projections of $i_1p_1+i_2p_2$ from $A\times B$ are respectively $p_1,p_2$ by bilinearity, so
$$
i_1p_1+i_2p_2=1_{A\times B}.
$$
For $A=B$, composing on the left by $\nabla_B$ and on the right by $\langle f,g\rangle$ forces the displayed formula for $f+g$. Thus \b[the biproducts determine exactly one such enrichment].
For the last part, the underlying one-object <category> has endomorphism <monoid> $(\mathbb N,\cdot)$, including $0$, and identity $1$. Its usual addition gives a <commutative-monoid enrichment>. Any permutation $\sigma$ of the <prime numbers> extends, by unique <prime factorization>, to a multiplicative <monoid automorphism> $\phi_\sigma:\mathbb N\to\mathbb N$, fixing $0,1$. Transport addition by
$$
\boxed{a\mathbin{\oplus_\sigma}b=\phi_\sigma^{-1}\bigl(\phi_\sigma(a)+\phi_\sigma(b)\bigr).}
$$
This is a <commutative monoid> operation with zero $0$, and multiplication distributes over it: applying $\phi_\sigma$ reduces each distributive law to the ordinary one in $\mathbb N$. Thus each operation supplies a <semi-additive structure> on the same fixed composition law.
For each <odd prime> $p$, take $\sigma$ to interchange $2$ and $p$, fixing every other prime. Then
$$
1\mathbin{\oplus_\sigma}1=\phi_\sigma^{-1}(2)=p.
$$
Different choices of $p$ give different additions, and there are infinitely many <prime numbers>. Hence there are \b[infinitely many distinct semi-additive structures], even though these transported structures are isomorphic as enriched <categories>. The underlying one-object <category> has no <terminal object>, because its endomorphism <set> is not a singleton, so it indeed has no finite <products in a category>.
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