= Solution
A <preadditive category> has an abelian-group structure on every <hom-set> and bilinear composition. One equivalent definition of an <additive category> is a <preadditive category> with a <zero object> and finite <products in a category>; the following argument proves that these <products in a category> are automatically <coproducts in a category>. An <abelian category> is an <additive category> with kernels and cokernels in which every <monomorphism> is a kernel and every <epimorphism> is a cokernel. Equivalently, for every arrow the canonical map from its coimage to its image is invertible.
Let $P=\prod_{i=1}^nA_i$, with projections $p_i$. Define $\iota_i:A_i\to P$ by $p_j\iota_i=\delta_{ji}$, where the diagonal entry is the identity and the others are zero. Since its composites with all projections are identities,
$$
\sum_i\iota_ip_i=1_P.
$$
Given $f_i:A_i\to B$, the map $f=\sum_if_ip_i$ satisfies $f\iota_i=f_i$. If another map has those composites, multiplying the displayed identity by that map forces it to equal $f$. Thus $P$ is also a <coproduct in a category>. The empty case is the <zero object>. Reversing arrows proves that a finite <coproduct in a category> is likewise a <product in a category>. Hence \b[finite <products in a category> and <coproducts in a category> are canonically <biproducts>] in an <additive category>.
For the <category of abelian groups>, let $(A_i,D_{ij})$ be a directed diagram, with no injectivity assumption on its transition maps. Its set <colimit> consists of classes $[i,a]$, with
$$
[i,a]=[j,b]\quad\Longleftrightarrow\quad D_{ik}a=D_{jk}b\text{ for some }k\geq i,j.
$$
Directedness proves transitivity of this relation. Define addition by moving representatives to a common upper index:
$$
[i,a]+[j,b]=[k,D_{ik}a+D_{jk}b].
$$
Moving all indices involved to one further upper bound proves independence of representatives and of $k$. Define $-[i,a]=[i,-a]$ and use the common class of the zeros as zero. The abelian-group laws hold at a common index, so descend to the quotient. Every canonical map $A_i\to C$ is a homomorphism, and a compatible cocone induces the homomorphism $[i,a]\mapsto f_i(a)$. This is the unique extension. Moreover the formula for addition is forced by those canonical homomorphisms, proving that the underlying-set <functor> creates directed <colimits>. This establishes <directed colimits created by the abelian-group forgetful functor> explicitly.
If a cocone $\nu_i:A_i\to B$ has injective legs, the induced map $C\to B$ is injective. Indeed, equal images of $[i,a]$ and $[j,b]$ give, at a common index $k$,
$$
\nu_k(D_{ik}a)=\nu_i(a)=\nu_j(b)=\nu_k(D_{jk}b).
$$
Injectivity of $\nu_k$ gives equal representatives at $k$. Since <monomorphisms> of <abelian groups> are exactly injective homomorphisms, we have proved that \b[the <category of abelian groups> is a finitary <abelian category>] in the directed-union sense of the question.
Now let $\mathcal A$ be complete, cocomplete and finitary. Index by finite subsets $J\subseteq I$ and put $S_J=\bigoplus_{j\in J}A_j$. For $J\subseteq K$, insert the additional zero coordinates. A compatible cocone on these finite <biproducts> is exactly a family of maps from every $A_i$, so their directed <colimit> is $S=\coprod_{i\in I}A_i$.
Each finite <biproduct> has a map $k_J:S_J\to P=\prod_{i\in I}A_i$ using its coordinates in $J$ and zeros outside $J$. Projection to the finite set of coordinates in $J$ is a retraction, so $k_J$ is a <split monomorphism>. These maps form a cocone of <monomorphisms>. The finitary property makes its induced map a <monic arrow>, and that induced map is exactly the canonical infinite identity matrix:
$$
\boxed{j:\coprod_{i\in I}A_i\longrightarrow\prod_{i\in I}A_i\text{ is monic}.}
$$
This proves the <coproduct-to-product comparison in a finitary abelian category>. The PDF's use of finite <products in a category> in the hint is valid because these are the same finite <biproducts>.
If $\mathcal A$ is also a <cofinitary abelian category>, apply this argument in its opposite <category>. The same canonical comparison $j$ is then an <epic arrow> in $\mathcal A$. In an <abelian category> an arrow that is both a <monomorphism> and an <epimorphism> is invertible, so $j$ is an isomorphism.
For countably many copies of $A$, put $S=\coprod_{n\geq0}A$ and let $q=j^{-1}\Delta:A\to S$, where $\Delta$ is the <product in a category> diagonal. Let $\iota_n$ denote the <coproduct in a category> injections, let $s:S\to S$ shift them by $s\iota_n=\iota_{n+1}$, and let $\nabla:S\to A$ be the codiagonal. Comparing every <product in a category> coordinate gives
$$
jq=j\iota_0+jsq.
$$
Since $j$ is a <monic arrow>, $q=\iota_0+sq$. Also $\nabla\iota_0=1_A$ and $\nabla s=\nabla$. Therefore, for $z=\nabla q$,
$$
z=1_A+z.
$$
<Hom-sets> are <abelian groups>, so cancellation gives $1_A=0$. Any arrow to or from $A$ is consequently zero by composition with its identity. Thus every object is both initial and terminal, and
$$
\boxed{\mathcal A\text{ is degenerate: every object is a zero object}.}
$$
This is the countable shift-and-fold argument behind <countable biproducts in an additive category force triviality>.
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