My instructor for abstract algebra explained that in order to show that $(G, \star)$ is a group over the set $G$ all we have to do is verify that the identity is in G and that for $a, b \in G$ we have $a\star b^{-1} \in G$
Is there a similarly compact way to verify that $(R, +, \times)$ is a ring over the set $R$? What about verifying that $M$ is a left (or right) $R-$module for a ring $R$?
I suspect the answer to this question will be more nuanced for modules since there is a conjugation map and a homomorphism happening but I am not sure for either modules or rings if there is a more compact method than independently verifying each part of the definitions.
What you are describing is a compact test for determining if a nonempty set is a subgroup of a group.
The proposed test does not guarantee associativity of the product, so it is incomplete.
However, when applied to a subgroup, associativity is inherited from the supergroup (which is a group by assumption), so it is not an issue.
The analogue for rings would be a subring test where you first check if $(S,+)< (R,+)$ using the criterion above ($a-b\in S$ for all $a,b\in S$) and then you make sure it is closed under multiplication ($ab\in S$ for $a,b\in S$.)
Perhaps more importantly, the latter can be extended to be test to see if a nonempty subset is an ideal if the last condition is replaced with $ar,ra\in S$ for every $a\in S$, and $r\in R$.
I think at this point you can probably work out the analogue for modules yourself.