Note that if $A_n$ is any family of sets, then
$$ \bigcup_{n\in\mathbb N}A_n = \bigcup_{n\in\mathbb N}\bigl(A_n\setminus\bigcup_{k=0}^{n-1} A_k\bigr)$$
where the summands on the right-hand side are disjoint, and each of them is constructed from finitely many of the $A_i$s by a sequence of complements and finite unions.
So if you choose you can restrict yourself to requiring countable unions of disjoint set, if you also require that the algebra is closed under finite unions of arbitrary sets.
A Dynkin system that is not a $\sigma$-algebra:
If you don't require arbitrary finite unions, what you get is not necessarily a $\sigma$-algebra. Consider for example the system of subsets of $\mathbb R$ consisting of $\varnothing$, $\{0,x\}$ for every $x\ne 0$, and the complements of these sets. It is closed under your proposed axioms, because the only nontrivial disjoint union there is to take is $\{0,x\} \cup \{0,x\}^\complement = \mathbb R$.
A small finite example is
$$ \bigl\{\varnothing, \{0,2\}, \{0,3\}, \{1,2\}, \{1, 3\}, \{0,1,2,3\} \bigr\} $$
Let $A$ and $B$ be sets in $S'$. Then there are finite disjoint collections $\{A_i\}_{i=1}^n$ and $\{B_j\}_{j=1}^m$of sets in $S$, such that $A= \bigcup_{i=1}^n A_i$ and
$B= \bigcup_{j=1}^m B_j$.
It is easy to see that $\{A_i\cap B_j\}_{i,j=1}^n$ is a finite disjoint collection of sets in $S$. So,
$$ A\cap B= \bigcup_{i,j=1}^n (A_i\cap B_j) \in S' \tag{1}$$
Now note that
$$A \cup B_1 = \left (\bigcup_{i=1}^n A_i \right )\cup B_1 = \left (\bigcup_{i=1}^n (A_i \setminus B_1) \right )\cup B_1 $$
For each $i$, $1\leqslant i \leqslant n$ there is a finite disjoint collection $\{D_{i,k}\}_{k=1}^{n_i}$ of set in $S$ such that $A_i \setminus B_1= \bigcup_{k=1}^{n_i}D_{i,k}$. Since $\{A_i\}_{i=1}^n$ is a finite disjoint collection of sets in $S$, we can conclude that the collection $\{ D_{i,k} : 1 \leqslant i \leqslant n, 1 \leqslant k \leqslant n_i \} \cup \{B_1\} $ is a finite disjoint collection of sets in $S$. So, we have
$$ A \cup B_1 = \left (\bigcup_{i=1}^n (A_i \setminus B_1) \right )\cup B_1 =
\left (\bigcup_{i=1}^n \bigcup_{k=1}^{n_i} D_{i,k} \right )\cup B_1 \in S'
$$
So, we have that $ A \cup B_1 \in S'$. By finite induction, we have that
$$ A\cup B = A \cup \left (\bigcup_{j=1}^m B_j \right ) =( ( ( (A\cup B_1)\cup B_2) \cdots )\cup B_m) \in S' \tag{2} $$
From $(1)$ and $(2)$, by finite induction, we have that $S'$ is closed with respect to the finite unions and finite intersections.
Remark: Since the question is about semirings, we must be careful about the term "finite intersection". If we take an empty collection of sets in $S'$, the intersection of such empty family is, by definition, $X$ which may not be in $S$ nor in $S'$. So, we understand here "finite intersection" as being the intersection of a non-empty family of sets in $S'$.
Best Answer
No. If $M$ is too small (i.e., empty) or too large (i.e., not a set) then it is not a $\sigma$-algebra. But those are trivial cases. Here is a less pathological counterexample:
Let $X=\{1,2,3,4\}$ and $$M=\big\{\emptyset,\;\;\{1,2\},\;\;\{2,3\},\;\;\{3,4\},\;\;\{1,4\},\;\;\{1,2,3,4\}\big\}$$ Then $M$ is a monotone class closed under complements and finite disjoint unions, but not under intersections.