Using your definition of "ring of sets in measure theory", that it is a (nonempty) collection of sets $R$ such that (1) it is closed under union ($\forall A,B\in R$ we have $A\cup B\in R$) and (2) it it is closed under set-theoretic difference ($\forall A,B\in R$ we have $A-B\in R$), we can in fact show $R$ is a commutative ring in the algebra sense, however with respect to the operations symmetric difference and intersection. So first we show that in fact, any ring of sets in measure theory is closed over symmetric difference and intersection:
Lemma. Let $R$ be a ring of set, then for all $A,B\in R$, we have $A\Delta B\in R$ and $A\cap B\in R$. (Here $A\Delta B:=(A-B)\cup(B-A)$ is the operation symmetric difference.)
Pf. The closure of symmetric difference is clear from its definition. And note well that $A\cap B= (A\cup B)-(A\Delta B)$. $\Box$
Now we make the observation: If $R$ is a ring of sets, then $\varnothing\in R$, since for any $A\in R$, we have $A\Delta A=\varnothing \in R$. Further note that for any $A\in R$, we have $\varnothing\Delta A=A$. Hence if we identify the operation $\Delta$ as "addition" and $\varnothing$ as the "additive identity", and every $A\in R$ is its own "additive inverse", we have
Claim. If $R$ is a ring of sets, then $(R,\Delta)$ is an abelian gorup. $\Box$ (Associativity given set-theoretically.)
Now, identify the operation $\cap$ as "multiplication", then with the following
Fact. Intersection distributes over symmetric difference. $\Box$
we have finally:
Proposition. If $R$ is a ring of set (measure theory sense), then it is also closed under symmetric difference $\Delta$ and intersection $\cap$. And that $(R,\Delta,\cap)$ forms a commutative ring (algebra sense). $\Box$
Let $\Omega=\{a,b,c\}$ and $\mathcal{C}=\{\Omega,\emptyset,\{a\},\{b\},\{c\}\}$ then $\mathcal{C}$ is a semi-algebra but NOT an algebra.
Best Answer
Here is a counterexample showing that 1,2, and 3 do not prove 3'.
Let $X$ be the nodes of an infinite complete binary tree. Then for $x\in X$, let $L(x)$ denote all nodes in the left subtree from $x$, and similarly let $R(x)$ denote all nodes in the right subtree from $x$. Then let
$S = \{\{x\}| x\in X\} \cup \{\{x\}\cup L(x)| x\in X\} \cup \{\{x\}\cup R(x)| x\in X\} \cup \{\{\} \}$
In other words, S is comprised of all singletons, all singletons with their left subtrees, and all singletons with their right subtrees. One can check that this is a semi-algebra in the sense of 1,2,and 3. But we will never be able to write $X$ (the complement of the empty set) as a finite disjoint union of elements of $S$.