A binomial ring is a ring (for the purposes of this question all rings are commutative and unital) which is torsion-free and has, for each $n$, a binomial function $\binom{x}{n}$ satisfying $n!\binom{x}{n} = x(x-1) \cdots (x-n+1)$. The category $\mathsf{BRing}$ of binomial rings is a full subcategory of the category $\mathsf{Ring}$ of rings. This paper shows (section 7) that the full subcategory $\mathsf{BRing} \subset \mathsf{Ring}$ is both reflective and coreflective. This implies that $\mathsf{BRing}$ is closed under limits and colimits in $\mathsf{Ring}$ (colimits exist because they can be formed in $\mathsf{Ring}$ and then reflected into $\mathsf{BRing}$, and they agree with colimits in $\mathsf{Ring}$ because the inclusion has a right adjoint, the coreflection. Dually for limits.)
But I just constructed a parallel pair of homomorphisms of binomial rings whose coequalizer is $\mathbb{Z}/2$, which is not torsion-free. So it looks like $\mathsf{BRing}$ is not closed under coequalizers. In fact, the coequalizer of these two maps in $\mathsf{BRing}$ exists, and is given by the zero ring. So the inclusion $\mathsf{BRing} \to \mathsf{Ring}$ doesn't preserve colimits, and can't have a right adjoint. Right?
Let me re-state the construction here, in hopes that someone will find a hole in it. Let $\mathcal{B}$ be the category of universal algebras whose operations are the ring operations, along with, for each $n$, a unary operation $\binom{-}{n}$, and whose relations are the ring equations, along with equations $n!\binom{x}{n} = x(x-1)\cdots (x-n+1)$. Then $\mathsf{BRing}$ is concretely isomorphic to a full subcategory of $\mathcal{B}$ (consisting of the algebras whose addition is torsion-free), since any ring homomorphism in $\mathsf{BRing}$ also preserves the $\binom{-}{n}$ operations by torsionfreeness.
Let $F[x,y]$ be the free object of $\mathcal{B}$ on two generators. Obviously $F[x,y]$ is torsion-free. There is a quotient map $q: F[x,y] \to \mathbb{Z}/2$, where $\mathbb{Z}/2$ is equipped with a $\mathcal{B}$-algebra structure with the usual underlying ring, and each $\binom{-}{n}$ operation identically equal to zero (actually, each $\binom{-}{n}$ can be defined arbitrarily, it doesn't matter). Say that $q(x)=0$ and $q(y)=1$. Then, since $\mathcal{B}$ is an algebraic category, the kernel pair of $q$ exists in $\mathcal{B}$ and is given by $K = \{(a,b) \mid q(a)=q(b)\} \subset F[x,y] \times F[x,y]$. As a subgroup of a product of torsionfree additive groups, the additive group of $K$ is torsionfree, so $K$ lies in $\mathsf{BRing}$.
So the two projections $K \stackrel{\to}{\to} F[x,y]$ form a coequalizer diagram in $\mathsf{BRing}$. Its coequalizer in $\mathsf{Ring}$ is $\mathbb{Z}/2$, because there are only two equivalence classes. But since $\mathbb{Z}/2$ is torsion, it doesn't lie in $\mathsf{BRing}$. So the inclusion $\mathsf{BRing} \to \mathsf{Ring}$ doesn't preserve this coequalizer.
Aha! the mistake comes, naturally enough, at the word "obviously". The free $\mathcal{B}$-algebra on a set is not torsion-free. For example, we have $2\binom{x}{2}(x-2) = x(x-1)(x-2) = 6 \binom{x}{3}$, but we don't have $\binom{x}{2}(x-2) = 3 \binom{x}{3}$. For example, by making $\mathbb{Z}/2$ a $\mathcal{B}$-algebra with $\binom{x}{2}=1$ and $\binom{x}{3}=0$, we violate this equation with $x=1$.