Closure properties can be formulated in terms of concepts from universal algebra. Let $X$ be the underlying set (in our examples, $X$ is a famly of sets itself). Let $I$ be an index set, $(\kappa_i)_{i\in I}$ be a family of cardinal numbers and $(f_i)_{i\in I}$ a family of function satisfying $f_i:X^{\kappa_i}\to X$ for all $i$. We say that $C\subseteq X$ is closed under $(f_i)_{i\in I}$ if we have for all $i\in I$ that $f_i(x)\in C$ for all $x\in C^{\kappa_i}$. One can show that the family of sets closed under $(f_i)_{i\in I}$ forms a Moore collection.
Let's look an an example: Let $U$ be a set and $X\subseteq 2^U$. We let $I=\{s,c,u\}$, $\kappa_s=0$, $\kappa_c=1$, and $\kappa_u=\omega$. We identify constants and nullary functions, so we can let $f_s=U$. We let $f_c(A)=A^C$ for all $A\in X$, and we let $f_u(A_0,A_1,\ldots)=\bigcup_n A_n$. That $X$ is closed under these three functions means simply that it contains $X$, is closed under complements and countable unions- it is a $\sigma$-algebra.
Now, one cannot write down semi-algebras this way, since there is no unique decomposition of the complement into disjoint sets. If $\mathcal{S}$ is a semi-algebra and $A\in\mathcal{S}$, then there exists a number $n$ and sets $B_1,\ldots,B_n\in\mathcal{S}$ that are disjoint and such that $A_c=B_1\cup\ldots\cup B_n$. Now if there exists a unique such family and if this family only depended on $A$, we could write down this property as closure under some functions in the following way: We let $f_{c_1}=B_1,\ldots, f_{c_n}=B_n$, and for $m>n$ we let $f_{c_m}=f_{c_n}$. We use the last condition because we have no a priori bound on how many sets are needed. But these sets are not a function of $A$, so this property can not be viewed as a closure property.
Here is an explicit example (taken from Alprantis & Border) that shows that the intersection of sem-algebras might fail to be a semi-algebra: Let $X=\{0,1,2\}$, $\mathcal{S}_1=\big\{\emptyset, X,\{0\},\{1\},\{2\}\big\}$, $\mathcal{S}_2=\big\{\emptyset, X,\{0\},\{1,2\}\big\}$, and $A=\{0\}$. We have $\mathcal{S}_1\cap\mathcal{S}_2=\big\{X,\emptyset,\{0\}\big\}$, and $A^C=\{0\}^C=\{1,2\}$ is not the disjoint union of elements of this intersection.
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). $\Box$
Best Answer
Always remember that the names we pick for things are a matter of convenience, and there are not really any rules to follow. (But it helps when people pick predictable things!)
Here's a fast answer that is probably not historically accurate, but will probably put your mind at ease: in universal algebra an algebra is just some set with different operations and rules acting in it. In that sense, groups, rings, rings of sets etc are all just generic "algebras". So you can see some people (at least) don't mind using "algebra" very flexibly.
As you have found out, the -set versions of rings and algebras are a little different from the algebraic ones. Let's focus on the similarities for a moment, to see why the names are kind of parallel to each other:
Ring and ring-of-sets: Both involve a set closed under two operations.
Field and field-of-sets: Both involve a set closed under two operations, plus a unary operation (multiplicative inverse/complementation)
The case of a "Boolean algebra" is interesting, because it kind of lies at the intersection of these two notions. While someone said they are lattice theoretic, it is also important to remember that they really are honest-to-goodness rings, too.
The use of "semi-" in front of terms has a pretty consistent use, and that is just to say that it is not quite as strong as the usual version. This is true for both a semi-ring-of-sets and a semiring.
To find an analogue of $\sigma$-algebras in ring theory, we would have to think of a field with infinitary operations; however, I don't know if anything like that exists. I do have an easy example of a semiring with infinitary operations, and that is the semiring of ideals of a ring. (That is, the set is the set of ideals of a fixed ring, along with the operations of ideal addition and ideal multiplication.)
For your example of a $\pi$-system, I think the best analogue is a semigroup, since there are no "inverses" provided by the complement. If you took a $\pi$-system and required it to be closed under complements, then I would be more inclined to analogize that to a group.