It is a standard fact that the module of Kähler differentials $\Omega_L$ of a finite separable extension of a field $k$ is equal to 0.
I also know that for a ring B and a finite extension $k \rightarrow B$ a sufficient condition for the module of differentials to vanish is that $B$ is a finite product of separable extensions of $k$.
Question.
Is it possible to lift this result to more general extensions $A\rightarrow B$ where $A$ and $B$ are both rings? Of course with some additional hypothesis on the rings and on the extension.
Best Answer
Let me answer the literal questions the OP asked, and then give a solution to the question he was really getting at (elucidated in the chat linked in the above comments).
The literal question
Let's suppose that $X\to Y$ is a finite flat map (let's assume that $Y$ is Noetherian to avoid headaches). Then, to check that $X\to Y$ has vanishing differentials is equivalent to checking that $X\to Y$ is etale. To do this without actually computing the differentials is, in general, not easy. There are a few ways one can do it in practice:
Of course, these might all be harder than just computing the differentials given what you know about $X\to Y$.
Another useful technique (which I implement below) is that to check that $X\to Y$ is etale it suffices to check this over some fpqc cover of $Y$. More concretely, since you like commutative algebra, if we have a ring map $A\to B$ and we have some faithfully flat map $A\to R$ (e.g. like $A\to\widehat{A}$ if $A$ is local) then $A\to B$ is etale if and only if $R\to B\otimes_A R$ is. This falls into the general notion of 'fpqc descent'. But, one concrete way of seeing it is that the non-degeneratness of the trace pairing (mentioned in the third bullet above) can be checked after base changing to a faithfully flat extension.
As for the question in the comments (paraphrasing):
The answer is no as $\mathrm{Spec}(\mathbb{Z}[i])\to\mathrm{Spec}(\mathbb{Z})$ shows. If you want them to be (even excellent!) DVRs you can consider $\mathrm{Spec}(\mathbb{Z}[i]_{(1+i)})\to\mathrm{Spec}(\mathbb{Z}_{(2)})$.
The question the OP was actually interested in
I will write here the actual question the OP was interested. If the OP decides to transfer it from our chat to a question on the main forum I will transfer this answer there.
My proof might be overly complicated. I'm sure there is some simpler proof just messing around with augmentation ideals. I also tried to write the below as commutative algebra-y as possible at the request of the OP--but I don't like commutative algebra, so it didn't end up optimally algebra-y.
Solution: Let us suppose first that $G$ is finite etale over $Y$. We can decompose $G$ into connected components $\displaystyle G=\bigsqcup_i \mathrm{Spec}(B_i)$. We claim that each $B_i$ is a domain. The point is that since $B_i$ is Noetherian we know that it suffices to show that all the local rings of $B_i$ at its maximal ideals are domains. Note though that since $A\to B_i$ is unramified we have for any maximal $\mathfrak{q}$ of $B_i$ the map $A\to B_\mathfrak{q}$ has the property that $\pi B_\mathfrak{q}$ is equal to $\mathfrak{q}$. This means that $B_\mathfrak{q}$ is a regular local ring (since $\dim B_\mathfrak{q}=1$ since it's finite over $A$) and thus a domain.
Remark: The above is a half-way proof to the general fact that the connected finite etale covers of an integral normal scheme are integral normal.
Suppose, conversely, that $\displaystyle G=\bigsqcup_i \mathrm{Spec}(B_i)$ with $B_i$ normal domains. Note that since $\dim B_i=1$ we see that normality implies that for each closed point $\mathfrak{q}$ of $B_i$ we have that $(B_i)_\mathfrak{q}$ is a DVR and so is its completion. Note then that we have that
$$H:=G_{\widehat{A}}\cong \bigsqcup_i \mathrm{Spec}(B_i\otimes_A \widehat{A})=\bigsqcup_i \bigsqcup_{\mathfrak{q}\in\mathrm{MaxSpec}(B_i)}\mathrm{Spec}(\widehat{(B_i)_{\mathfrak{q}}})$$
We deduce that $H$ has connected components which are normal domains.
But, by the usual connected-etale sequence we have that $H$ is etale over $\widehat{A}$ iff its connected component of the identity $H^\circ$ is trivial (e.g. see section 8.1 of this). Note though that by our above decomposition, all the connected components of $H$ are normal and integral. In particular, $H^\circ=\mathrm{Spec}(B_0)$ for some normal domain $B_0$. This then implies that its generic fiber $H_\eta^\circ:=\mathrm{Spec}(B_0\otimes_{\widehat{A}}K)$ (with $K:=\mathrm{Frac}(\widehat{A})$) is normal and integral.
But, since $H^\circ_\eta$ is finite over $\mathrm{Spec}(K)$ and integral we know that $H^\circ_\eta=\mathrm{Spec}(C)$ where $C$ is a field. Note though that we have an identity section $\mathrm{Spec}(K)\to H^\circ_\eta$ since $H^\circ_\eta$ is a group over $\mathrm{Spec}(K)$. This implies that we have a map $C\to K$ of $K$-algebras. Since $C$ is a field this must be an isomorphism. Thus, $C=K$. So we see that $B_0\otimes_{\widehat{A}} K\cong K$. Since $B_0$ is finite free over $\widehat{A}$ this implies that $B_0$ is rank $1$. The conclusion follows. $\blacksquare$