By “(possibly many-sorted) algebraic category”, I mean “category of models of a (possibly many sorted) Lawvere theory. I have arguments for both affirmative and negative answers, and I can’t find a hole in either of them. Let $\mathsf{TF}$ denote the category of torsionfree abelian groups.
For background, though, note that $\mathsf{TF}$ is locally finitely presentable: it admits a finite-limit sketch given by adjoining to the sketch for abelian groups with carrier object $X$ the condition that the equalizer of $0, n: X \overset{\to}{\to} X$ is $0$ (where $0$ is the zero map, and $n$ is the multiply-by-$n$ map).
Here’s an abstract argument that $\mathsf{TF}$ is algebraic. $\mathsf{TF}$ can be sketched by a finite-product / coproduct sketch (EDIT: Zhen Lin points out that this is the mistake!): adjoin to the sketch for abelian groups with carrier object $X$ an object $Y$ and a coproduct diagram $X^0 \overset{0}{\to} X \overset{i}{\leftarrow} Y$, so that $Y$ consists of the nonzero elements of the group; adjoin also arrows $Y \overset{n}{\to} Y$ along with commuting squares of the form
$\matrix{X & \overset{n}{\to} & X \\
i \uparrow & & i \uparrow \\
Y & \overset{n}{\to} & Y}$
where $i: Y \to X$ is the coproduct injection, $n: Y \to Y$ is the new map, and $n: X \to X$ is multiplication by $n$.
This paper shows (Lemma 3.1) that any category sketchable by a finite-product / coproduct sketch is multialgebraic, so $\mathsf{TF}$ is multialgebraic. And since any cocomplete multialgebraic category is algebraic (a cocomplete category has multicolimits, this paper shows (Theorem 4.4) that a multicocomplete multialgebraic category is a generalized variety and (Theorem 3.10) that a cocomplete generalized variety is a variety, where here “variety” means “algebraic category”), it follows that $\mathsf{TF}$ is algebraic.
Here’s a concrete argument that $\mathsf{TF}$ is not algebraic. The finitely-presentable objects of $\mathsf{TF}$ are just the finitely-presentable groups which are torsion-free, i.e. the finitely-generated free abelian groups. This is because $\mathsf{TF}$ is closed in $\mathsf{Ab}$ under filtered colimits, so the same argument as in $\mathsf{Ab}$ shows that the finitely-presentable objects of $\mathsf{TF}$ are finitely-generated in the usual sense (every object $X$ is a filtered colimit of its finitely-generated subobjects, which are torsion-free if $X$ is, so $\mathrm{id}_X$ must factor through one of these subobjects if $\mathrm{Hom}(X,-)$ preserves this filtered colimit).
The Lawvere theory of an algebraic theory can be recovered (up to Cauchy completion) as the category of strongly finitely-presentable objects (objects $X$ such that $\mathrm{Hom}(X,-)$ preserves sifted colimits, i.e. preserves filtered colimits and reflexive coequalizers), which is a full subcategory of the finitely-presentable objects, closed under finite coproducts and splitting of idempotents. But the only such subcategories of the finitely-generated free abelian groups are (a.) the category consisting of the zero group and (b.) all the finitely-generated free abelian groups. But in the former case, the associated category of algebras is the terminal category while in the latter case it is just $\mathsf{Ab}$ — so we can’t get $\mathsf{TF}$.
So I’m left scratching my head. Which conclusion is correct? Is either argument correct? The step that feels like magic to me is the deduction that the category of models of a finite-product / coproduct sketch is multialgebraic (after all, a finitely sketchable category need not be finitely accessible), but it looks pretty straightforward in the linked paper.
I should mention that this question grew out of my attempt to answer a question of Martin Brandenburg, which is related, and which I’m still interested in resolving.
EDIT So in light of Zhen Lin’s comment-answer, what we can conclude is that the category of torsionfree abelian groups and all homormophisms is locally finitely presentable (by the first sketch), and the category of torsionfree abelian groups and injective homomorphisms is multialgebraic (by the second sketch) — and hence it’s a generalized variety. But the former category is not algebraic, and the latter category is not complete.