First let me say something about the unital case. If $R$ is a unital commutative ring and $L(R)$ is finite, then $R$ since Artinian+commutative we have $R\cong R_1\times\cdots\times R_k$ for some local rings $R_i$. Ideals of a product are product ideals, so $L(R)\cong L(R_i)\times \cdots\times L(R_k)$. A local ring has a unique maximal ideal, so each $L(R_i)$ has a unique coatom. Since lattices with unique coatom are directly indecomposable, it follows that the direct factors of $L(R)$ are exactly the lattices $L(R_i)$, so we can phrase this observation as "Each direct factor of the lattice $L(R)$ has unique coatom".
This is a nontrivial lattice-theoretic restriction on $L(R)$. There does exist a five-element directly indecomposable lattice that does not have a unique coatom, and from this we may deduce that it is impossible for $L(R) = \{R, A, B, A\cap B, 0\}$ where $A$ and $B$ are incomparable and $A\cap B\neq 0$. This provides an example of a five-element distributive lattice that is not isomorphic to the lattice of ideals of a unital commutative ring.
Now let me say something about the nonunital case.
Let $M_n$ be the lattice of height two with $n$ elements of height one. Each $M_n$ is modular. If $M_n$ is isomorphic to the lattice of ideals of a unital ring, then from the previous paragraph we must have $n=1$ or $2$. Consider the nonunital case where $L(R)\cong M_n$, $n\geq 3$. The sum of any two distinct maximal ideals equals $R$ (always), and by the shape of $M_n$ the intersection of any two maximal ideals is zero. If $A, B, C$ are distinct maximal ideals, then $AR=A(B+C)=AB+AC\subseteq (A\cap B)+(A\cap C)=0$. Similarly, $BR=0$, so $RR=(A+B)R=AR+BR=0$. This says that $R$ has zero multiplication. It follows that the ideal lattice of $R$ coincides with its lattice of additive subgroups of $R$. It is not hard to show that if an abelian group $G$ has a lattice of subgroups isomorphic to $M_n$, then $n=p+1$ for some prime and $G\cong \mathbb Z_p\times \mathbb Z_p$. Putting this all together, we get that if
$M_n$ is isomorphic to the lattice of ideals of a not-necessarily-unital ring, then $n=1$ or $2$ or else $n = p+1$ for some prime $p$.
It follows from the two paragraphs above that $M_3$ is not isomorphic to the lattice of ideals of any unital commutative ring, and the lattice $M_5$ is not isomorphic to the lattice of ideals of any commutative ring.
In the question it is claimed that the distributive case follows from Birkhoff's Theorem by an easy argument. There are problems with the argument. First, in what way do the characteristic functions associated to a ring of sets form a ring? If your ring of sets contains only $A \subsetneq B\subsetneq C\subsetneq D$, then the characteristic functions of these sets are only $\chi_A, \chi_B, \chi_C, \chi_D$. What is $\chi_B+\chi_C$? Also, if you intend the characteristic functions to form a ring under pointwise multiplication (so $\chi_A\cdot \chi_B=\chi_{A\cap B}$), then the ring will be unital and Boolean. In this case $L(R)$ is a Boolean lattice. In particular, this construction will not represent distributive lattices that are not Boolean.
The problem statement contains the claim that $L(R_{M_3})\cong M_3$. This is not true, since $L(R_{M_3})$ has six elements and $M_3$ has five elements. (Moreover, $R_{M_3}$ has a unit, so $L(R_{M_3})\cong M_3$ is impossible by the third paragraph of this answer.)
There is speculation in the problem statement that "if a finite modular lattice appears as $L(R)$, then it does so for a ring of characteristic $2$". From the second paragraph of this answer, if $p$ is prime and $L(R)\cong M_{p+1}$, then $R$ must be of characteristic $p$.
