Let $\newcommand{\TT}{\mathbb{T}}\newcommand{\cAb}{\mathsf{Ab}}\TT_\cAb$ be the Lawvere theory of Abelian groups. It is a monoid in the category of Lawvere theories (the tensor product is given by some sort of "commutative coproduct": we put together freely all the operations of our two theories and we require that they commute). It's actually even an idempotent monoid, in the sense that the product $\TT_{\cAb}⊗\TT_\cAb→\TT_\cAb$ is an isomorphism. This is because of the Eckmann-Hilton argument (which shows also that the tensor product of two copies of the theory of groups is the theory of abelian groups).
Then a Lawvere theory has a category of algebras which is abelian if and only if it is equipped with a (necessarily unique) $\TT_\cAb$-action. In concrete terms, we need to have an interpretation of the theory of abelian groups in our algebraic theory in such a way that all the polynomials of the theory are linear. Linear combinations are themselves morphisms in the category of algebras and thus morphisms are stable by linear combinations.
Such an interpretation, if it exists, is unique. By the Eckmann-Hilton argument, we only need to show that the two zeros of two potential interpretations coincide. This is true because the zero of an algebra $X$ is given by the unique morphism from the terminal algebra to $X$.
Let us show more formally the above.
First, suppose that the category of algebras of $\TT$ is abelian. Then there is an interpetation of the theory of abelian groups in $\TT$ because for instance $(x,y) ↦ x+y$ corresponds to the morphism $\newcommand{\Free}{\operatorname{Free}}\Free(1)→\Free(1)⊔\Free(1)$ given by $i_1+i_2$ with $i_1$ and $i_2$ the two canonical injections. Let's see why this operation of the theory commutes with all the other operations. This means that for all $\TT$-algebra $X$, the sum defines a morphism $X^2→X$. This function is given by precomposition $\newcommand{\Hom}{\operatorname{Hom}}\Hom(\Free(2),X) → \Hom(\Free(1),X)$ with $i_1+i_2$. But we have $(i_1+i_2)f = i_1f + i_2f$ for all $f ∈ \Hom(\Free(2),X)$, so this function $X^2→X$ is actually $p_1+p_2$ with $p_1$ and $p_2$ the two canonical projections. The same goes for any linear combination instead of the sum $x+y$.
Reciprocally, we need to show that if there is an interpretation of the theory of abelian groups in $\TT$ commuting with all the other operations of $\TT$, then the category of $\TT$-algebras is abelian. We already saw why it is enriched over $\cAb$. We also see that $X×Y$ is the coproduct of $X$ and $Y$: if we have two morphisms $f:X→Z$ and $g:Y→Z$, we have the morphism $(x,y) ∈ X×Y ↦ f(x)+g(y)$ (given as the composite $X×Y→Z×Z→Z$ so we see it's a morphism). It is the unique morphism $X×Y→Z$ restricting to $f$ and $g$ along the two injections $x↦(x,0)$ and $y↦(0,y)$. The category of $\TT$-algebras has kernel and cokernel since it has all limits and colimits. And the image is given by the usual image, so it's abelian.
Here is an explanation of the link of this characterization with the one of Vladimir Sotirov.
An action of $\newcommand{\TT}{\mathbb{T}}\newcommand{\Ab}{\mathbf{Ab}}\TT_\Ab$ on $\TT$ is a morphism $α:\TT_\Ab⊗\TT→\TT$ satisfying some axioms. But $\TT_\Ab$ is idempotent and these axioms reduce to the fact that the canonical map $\TT→\TT_\Ab⊗\TT→\TT$ is the identity (this is a general fact of idempotent monads). There is a morphism $\TT_\Ab→[\TT,\TT]$ corresponding to $α$, which means that we have an abelian group structure on an object of $[\TT,\TT]$, and the axiom of the previous sentence says that it is the identity. It is also true in general that an abelian group is a commutative monoid $X$ such that $+,π_1 : X^2 → X$ are a product structure, so we see that the two characterizations are equivalent.