Freyd's axioms force the additive structure and recover the AB2 definition

Freyd's axioms already make $\mathcal{A}$ balanced: if $m:A\to B$ is monic and epic, the normality clause writes $m$ as a kernel of some $g:B\to C$, so $gm=0$; because $m$ is epic, $g=0$, and the identity $1_B$ is a kernel of $0_{B,C}$, so $m$ is isomorphic to $1_B$.

Let $S=A\bigsqcup B$ and $P=A\times B$ be the coproduct and product supplied by [L1]. The split epics $[1_A,0]:S\to A$ and $[0,1_B]:S\to B$ are cokernels of the opposite injections because maps out of a coproduct are determined by the injections; dually the split monics $⟨1_A,0⟩:A\to P$ and $⟨0,1_B⟩:B\to P$ are kernels of the opposite projections because maps into a product are determined by the projections. Therefore the canonical comparison $c:S\to P$ is both monic and epic, so step 1.1 makes it an isomorphism. Thus binary biproducts exist, and together with the zero-object clause this gives finite biproducts.

By [L3], the finite biproducts from step 2.1 give a canonical commutative-monoid law on each hom-set. On $A\oplus A$, Mitchell's shear $\theta =\left(\begin{array}{cc}{1}_A& 1_A\\ 0& 1_A\end{array}\right)$ is monic and epic by the same kernel-cokernel argument used in step 2.1, hence invertible by step 1.1. Writing ${\theta }^{-1}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, the matrix identity $\theta {\theta }^{-1}=1$ yields $1_A+b=0$ in the monoid law of [L3]. For every $x:A\to B$, the morphism $xb$ is therefore an additive inverse of $x$, so the hom-monoids are abelian groups. Hence $\mathcal{A}$ is preadditive, and with finite biproducts it is additive.

Let $f:A\to B$, let $k:K\to A$ be its kernel, let $p:A\to Q=\mathrm{coim}(f)$ be the cokernel of $k$, let $m:I=\mathrm{im}(f)\to B$ be the kernel of a cokernel of $f$, and let $u:Q\to I$ be the canonical morphism from [L4]. Put $i:=mu:Q\to B$. To show that $i$ is monic, let $x:X\to Q$ satisfy $ix=0$, let $q:Q\to Q^{\mathrm{\prime }}$ be a cokernel of $x$, and write $i=jq$. Since $qp$ is a composite of cokernels, [L5] makes it epic, so the conormality clause gives some $h:H\to A$ with $qp=\mathrm{coker}(h)$. Now $fh=iph=jqph=0$, so $h$ factors through $k$; hence $ph=0$. Because $qp$ is a cokernel of $h$, the map $p$ factors through $qp$ as $p=p^{\mathrm{\prime }}qp$. Since $p$ is epic by [L5], $p^{\mathrm{\prime }}q=1_Q$, so $q$ is monic. Then $qx=0$ forces $x=0$, and $i$ is monic.

The dual argument shows that the factorization map $e:=up:A\to I$ is epic: starting from a kernel of a map out of $I$, one passes to the opposite category and repeats step 4.1. Since $m$ is monic and $p$ is epic by [L5], the equalities $mu=i$ and $up=e$ imply that $u$ itself is monic and epic. Step 1.1 then makes $u$ an isomorphism.

Thus every morphism has kernels and cokernels and an invertible canonical coimage-to-image comparison. Together with the additivity from step 3.1, this is exactly the working abelian definition of Abelian category.