To which extent all infinite commutative unital rings $R$
with the following property have been classified?
Every $R$-module cannot be equal to a union of finite number of its proper submodules.
I will classify these rings without assuming commutativity. I claim
Thm. The following are equivalent for a ring $R$.
There is an infinite $R$-module $M$ that is the union of finitely many proper submodules.
$R$ has a finite nontrivial module.
$R$ has a proper ideal of finite index.
$R$ has a nontrivial finite quotient ring.
[2 implies 3] If $M$ is a finite nontrivial left $R$-module, then its left annihilator is a proper ideal of finite index.
[3 implies 4] If $I$ is a proper ideal of finite index in $R$, then $R/I$ is a nontrivial finite quotient ring.
[4 implies 1] Suppose that $S=R/I$ is a finite nontrivial quotient of $R$. By restriction of scalars, $S\oplus S$ is a (non-cylic) left $R$-module, so $S\oplus S = M_1+\cdots + M_n$ for some finite $n$ and for proper (cyclic) submodules $M_i\leq S\oplus S$. Now the infinite module $(S\oplus S)\oplus S^{\omega}$ can be written as the finite sum $\sum_{i=1}^n(M_i\oplus S^{\omega})$ of proper submodules. \\\
[1 implies 2] Suppose that $R$ has an infinite module $M$ that is a finite union of proper submodules, $M=\sum_{i=1}^n M_i$. I now employ an old and famous result of B. H. Neumann which asserts that if a group $G$ is a union of finitely many cosets of subgroups, say $G = \cup_{i=1}^n a_iH_i$, then one can discard all cosets of subgroups of infinite index and the remaining union still covers $G$. In particular, since modules have underlying group structure, the representation $M=\sum_{i=1}^n M_i$ forces some $M_i$ to have finite group-index in $M$. That is, $M/M_i$ is a finite nontrivial $R$-module. \\\
Neumann, B. H. Groups covered by permutable subsets. J. London Math. Soc. 29, (1954). 236–248.