Are differential rings monoids in a monoidal category? $\newcommand{\pt}{\mathrm{pt}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}$A number of algebraic structures can be defined as monoids in some appropriate monoidal category:

*

*A monoid                          is a monoid in $(\mathsf{Sets},\times,\pt)$;

*A semiring                        is a monoid in $(\mathsf{CMon},\otimes_{\N},\N)$;

*A ring                            is a monoid in $(\mathsf{Ab},\otimes_\Z,\Z)$;

*An $R$-algebra                    is a monoid in $(\mathsf{Mod}_R,\otimes_R,R)$;

*A graded $R$-algebra              is a monoid in $(\mathsf{Gr}_\Z\mathsf{Mod}_R,\otimes_R,R)$;

*A differential graded $R$-algebra is a monoid in $(\mathsf{Ch}_\bullet(\mathsf{Mod}_R),\otimes_R,R)$.

Is this also the case for differential rings?
 A: $\newcommand{\defeq}{\overset{\mathrm{def}}{=}}\newcommand{\id}{\mathrm{id}}\newcommand{\Mod}{\mathrm{Mod}}\newcommand{\pt}{\mathrm{pt}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\d}{\mathrm{d}}\newcommand{\dAb}{\mathsf{End}(\mathsf{Ab})}$DGAs are monoids in chain complexes. To get differential rings as monoids in some monoidal category, it suffices to remove the grading and the $\d^{2}=0$ condition.
In detail, consider the category $\mathsf{End}(\mathsf{Ab})\defeq\mathsf{Fun}(\mathbf{B}\N,\mathsf{Ab})$ whose

*

*Objects are pairs $(A,\d)$ with $A$ an abelian group and $d\colon A\to A$ a morphism of abelian groups.

*Morphisms $(A,\d_A)\to(B,\d_B)$ are morphisms of abelian groups preserving the derivation, i.e. such that the diagram
$$
\require{AMScd}
\begin{CD}
A @>\d_A>> A\\
@V f V V @VV f V\\
B @>>\d_B> B
\end{CD}
$$
commutes.

We can then put a monoidal structure $\otimes_\Z$ on $\dAb$ by defining
$$(A,\d_A)\otimes(B,\d_B)=(A\otimes_\Z B,\d_A\otimes_\Z1_B+1_A\otimes_\Z\d_B),$$
where the unit is given by the pair $(\Z,\d_\Z)$ with $\d_\Z\overset{\mathrm{def}}{=} 0$. Note that a morphism in $\dAb$ from $(\Z,\d_\Z)$ to $(A,\d_A)$ is just a "constant" element of $A$, i.e. an element with $\d_A a = 0$.
A monoid in $(\dAb,\otimes_\Z,(\Z,\d_\Z))$ will then be a triple $((A,\d),\mu,\eta)$ with

*

*$(A,\d)$ an object of $\dAb$; this accounts for the underlying additive abelian group of a differential ring and the derivation $\d$, which is $\Z$-linear;

*$\mu\colon(A,\d_A)\otimes_\Z(A,\d_A)\to(A,\d_A)$ a morphism of $\dAb$; this accounts for the multipication and the Leibniz rule: asking for the diagram
$$
\require{AMScd}
\begin{CD}
A\otimes_\Z A @>\d_A\otimes_\Z1_A+1_A\otimes_\Z\d_A>> A\otimes_\Z A\\
@V \mu V V @VV \mu V\\
A @>>\d_A> A
\end{CD}
$$
to commute is equivalent to asking
$$\d(ab)=\d(a)b+a\d(b)$$
to hold for all $a,b\in A$;

*$\eta\colon(\Z,\d_\Z)\to(A,\d_A)$ a morphism of $\dAb$, determining an element $1_A$ of $A$;

such that the usual associativity and unitality diagrams commute, which makes $(A,\mu,\eta)$ into a ring, and together with $\d$, this makes the quadruple $((A,\d),\mu,\eta)$ into a differential ring.
