= Solution
Use the following <group axioms> in the given language:
$$
\forall x\forall y\forall z\ ((x+y)+z=x+(y+z)),\qquad
\forall x\forall y\ (x+y=y+x),
$$
$$
\forall x\ (x+0=x),\qquad \forall x\exists y\ (x+y=0).
$$
<Commutativity> supplies the other identity and inverse equations. These <first-order sentences> therefore specify precisely an <abelian group> with the named identity $0$; an inverse <function symbol> is unnecessary because inverses are asserted to exist by a <quantified formula>.
For each positive <integer> $n$, let $t_n(x)$ be the <first-order term> recursively defined by $t_1(x)=x$ and $t_{n+1}(x)=t_n(x)+x$. Add the <torsion-free abelian group axiom scheme>
$$
\boxed{\forall x\ (t_n(x)=0\to x=0)\qquad(n=1,2,3,\ldots).}
$$
Every <first-order model> of these <first-order sentences> is an <abelian group> with no nonidentity element of <finite order>. Conversely every <torsion-free abelian group> satisfies each of the displayed <first-order sentences>. This is a countably infinite first-order axiomatization of exactly the required class.
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