If a ring $A$ is a $D$-module. And If $A\cong B$ as rings. Then is $A\cong B$ as $D$-module?
It seems obvious to me that yes, since everything that can be said about $A$ as a ring can be said about $B$ (also as a ring). Even so, I ask the question, because I have already come across some things that seemed "obvious" to me, however after reading the answer I realized that there was really some more restriction missing.
First...
If $A$ is a semigroup/group/ring/field/$R$-module/$k$-algebra/etc, and $f\colon A\to B$ is a bijection, then you can endow $B$ with a semigroup/group/ring/field/$R$-module/$k$-algebra/etc structure by using transport of structure. This seems to be what your "more 'elementary' question" in the comments are groping for. So, indeed, if $A$ is a ring and $f\colon A\to B$ is a bijection, then you can turn $B$ into a ring in a way that will make $f$ an isomorphism, by defining the operations on $B$ by $$\begin{align*} b_1+b_2 &= f(f^{-1}(b_1)+_A f^{-1}(b_2))\\ b_1\cdot b_2 &= f(f^{-1}(b_1)\cdot_A f^{-1}(b_2)), \end{align*}$$ where $+_A$ and $\cdot_A$ are the ring operations on $A$.
Likewise, if $A$ and $B$ are rings/abelian groups, $f\colon A\to B$ is an isomorphism of rings/abelian groups, and $A$ is endowed with a $D$-module structure, then you can define a $D$-module structure on $B$ by letting $$\alpha b = f(\alpha f^{-1}(b))\text{ for all }\alpha\in D, b\in B.$$ It is then easy to verify that this gives a $D$-module structure on $B$, and that $f$ will be a $D$-module isomorphism from $A$ to $B$.
However, it is not true in general that if $A$ and $B$ are rings that are $D$-modules, and we have a ring isomorphism $f\colon A\to B$, then $A$ and $B$ will be isomorphic as $D$-modules. It will be for some trivial cases; e.g., any ring can be viewed as a $\mathbb{Z}$-module by taking its underlying abelian group additive structure, and then the isomorphism of $A$ and $B$ will necessarily give you a $\mathbb{Z}$-module isomorphism as well; but for arbitrary rings $D$ this need not hold.
For example, take the ring $R=\mathsf{M}_{2\times 2}(k)$ of $2\times 2$\matrices with coefficients in a field $k$; viewed as a vector space over $k$, any linear transformation $T\colon \mathsf{M}_{2\times 2}(k)\to\mathsf{M}_{2\times 2}$ endows $R$ with a $k[t]$-module structure, by letting $p(t)$ act like the linear transformation $p(T)$. So you can take $A=B$. Now let $T_1$ be a diagonalizable linear transformation, and let $T_2$ be a linear transformation that makes $B$ irreducible (e.g., one with a single Jordan block). Then we can make $A$ into a $k[t]$-module by letting $t$ act like $T_1$, which will allow you to decompose $A$ into a direct sum of four submodules, whereas giving $B$ the $k[t]$-structure obtained by letting $t$ act like $T_2$ will not allow a similar decomposition. So the $k[t]$-module structures of $A$ and $B$ are not isomorphic, even though $A$ and $B$ are isomorphic as rings.