Category of binomial rings
A binomial ring is a commutative ring $R$ such that (1) the additive group of $R$ is torsionfree and (2) $n!$ divides $x(x-1)\dotsc(x-n+1)$ for all $n \in \mathbb{N}$ and $x \in R$. We may then define $\binom{x}{n} := \frac{x(x-1)\dotsc(x-n+1)}{n!}$ as usual. Jesse Elliott has proven here that a commutative ring is a quotient of a binomial ring if and only if (2) is satisfied. He calls these rings quasi-binomial.
I think that the concrete category $\mathsf{BRing}$ of binomial rings is not algebraic, i.e. up to equivalence the category of models of an algebraic theory. Likewise, the larger category $\mathsf{QBRing}$ of quasi-binomial rings is probably not algebraic. But how to prove this? Both categories are complete and cocomplete.
Solution 1:
$\mathsf{BRing}$ is algebraic, believe it or not! All the hard work to show this is done in Jesse Elliott's paper.
In fact, $\mathsf{BRing}$ is reflective and coreflective in $\mathsf{Ring}$ according to the paper (section 7), and it follows that it is closed in $\mathsf{Ring}$ under limits and colimits.
The paper also shows that the obvious forgetful functor $\mathsf{BRing} \to \mathsf{Set}$ has a left adjoint, sending a set $X$ to the ring of integer-valued polynomials with variables in $X$.
So we can apply the monadicity theorem: the forgetful functor $U: \mathsf{BRing} \to \mathsf{Set}$ has a left adjoint and reflects isomorphisms, so we just need to show it creates coequalizers of $U$-split pairs. This is true because the coequalizer of a $U$-split pair exists in $\mathsf{Ring}$, and $\mathsf{BRing}$ is closed under coequalizers.
This stands in stark contrast to torsion-free abelian groups, or torsion-free rings, both of which fail to be exact because they don't contain the quotient $\mathbb{Z} \to \mathbb{Z}/n$. It's actually very remarkable that $\mathsf{BRing}$ is closed under coequalizers in $\mathsf{Ring}$, and that no torsion can be created modding out by a binomially-closed congruence!
To be explicit: an operation in this algebraic theory is given by an integer-valued polynomial. Generators and relations for this can be found in the paper. It is not the theory freely generated by ring operations and operations $\binom{-}{n}$ satisfying $n!\binom{x}{n} = x(x-1)\cdots (x-n+1)$: there are further relations (I'm not sure if these are just the "obvious" ones like the one mentioned in item 2 below).
To document my wrong answers so far:
I attempted to give a finite-product - coproduct sketch for $\mathsf{BRing}$ to show it was multialgebraic and hence algebraic, but ended up sketching binomial rings and injective homomorphisms, a different category. Thanks to Zhen Lin for pointing this out!
I attempted to show that the strongly finitely presentable objects of $\mathsf{BRing}$ were the same as those in the full supercategory $\operatorname{Mod}(\mathcal{BR})$ of rings equipped with unary operations $\binom{-}{n}$ subject to $n!\binom{x}{n} = x(x-1)\cdots (x-n+1)$, which would mean $\mathsf{BRing}$ was not the category of models of the only possible Lawvere theory, and hence not algebraic. It turns out that this is wrong because the free objects of $\operatorname{Mod}(\mathcal{BR})$ don't even lie in $\mathsf{BRing}$ -- they have torsion! For instance, in $\operatorname{Mod}(\mathcal{BR})$, the equation $2\binom{x}{2}(x-2) = 6 \binom{x}{3}$ holds, but $\binom{x}{2}(x-2) \neq 3\binom{x}{3}$ does not, as evidenced by $\mathbb{Z}/2$, where the $\binom{-}{n}$ functions can be arbitrary and it's easy to violate this equation.
I attempted to show that $\mathsf{BRing}$ was not exact and hence not algebraic, by exhibiting a certain congruence with an exact quotient in $\operatorname{Mod}(\mathcal{BR})$ but not in $\mathsf{BRing}$. This was wrong because the quotient was of the free algebra in $\operatorname{Mod}(\mathcal{BR})$ on two generators, which doesn't lie in $\mathsf{BRing}$. In fact, since exactness is formulated entirely in terms of limits and colimits, the fact that $\mathsf{BRing}$ is closed in $\mathsf{Ring}$ under limits and colimits immediately implies that $\mathsf{BRing}$ inherits exactness from $\mathsf{Ring}$. More about this at a question I asked about it.
Update
Martin quite sensibly asks for generators and relations in the comments. After asking this question (and Martin's answer!), I believe that a full set of relations is given by the (special) $\lambda$-ring relations (with $\lambda^n(x) = \binom{x}{n}$), along with the identity shown by Darij Grinberg (with a slight typo; see his subsequent comment for a link to a proof) $\binom{x}{a}\binom{x}{b} = \sum_{i=a}^{a+b} \binom{i}{a}\binom{a}{a+b-i}\binom{x}{i}$, and of course the identity $n!\binom{x}{n} = x(x-1)\cdots (x-n+1)$. One can check, for example, that the relation $(x-n)\binom{x}{n} = (n+1)\binom{x}{n+1}$, as alluded to in item (2) above, follows from these relations.
Update Again Here's an argument that there's no 2-torsion: if $2x=0$, then $0=\binom{2x}{2}$ (in fact, the binomial operation applied to any integer must yield the usual result, being determined inductively by the addition rule for $\lambda$-rings (the Vandermonde identity)). This can be expanded using the multiplication rule as $4\binom{x}{2} + x$, and the first term is 0 using the rule $2\binom{x}{2} = x(x-1)$. With close study of these relations, I imagine such an analysis could be extended to show there's no torsion at all.