Let $a,b,c,d \in \mathbb N$, where $\mathbb N = \{0,1,2,\ldots\}$. Then we have $$ a + \mathbb N \cdot b \subseteq c + \mathbb N \cdot d $$ iff $a - c$ and $b$ are dividable by $d$.
Proof. Suppose we have $a + \mathbb N \cdot b \subseteq c + \mathbb N \cdot d$, then $a - c \in \mathbb N\cdot d$, hence $d$ divides $a-c$. Then choose some $n$ not dividable by $d$ and the equality $a - c + n\cdot b \in \mathbb N\cdot d$ implies that $b$ must be dividable by $d$. $\square$
So now generalize this in the following manner, if we have $2n$ numbers $c_1, \ldots, c_n$ and $d_1,\ldots, d_n$. Exactly when do we have $$ a + \mathbb N \cdot b \subseteq \bigcup_{i=1}^n c_i + \mathbb N \cdot d_i. $$ and the selection of those numbers is minimal in the sense that if we take some $c_j, d_j$ away, then $a + \mathbb N\cdot b$ is no longer in the union of the remaining ones, i.e. every number is mandatory. Is there any easy criterion like the one above in terms of these numbers?
I don't understand what did you mean with "easy criterion" but when we have $2n$ numbers $c_1, \dots, c_n$ and $d_1, \dots, d_n$ and $$ a+ \mathbb{N}\cdot b \subseteq \cup_{i=1}^{n}c_i+ \mathbb{N}\cdot d_i$$ and then we add 1 more $(c_{n+1},d_{n+1})$ which doesn't "relate" to $a$ and $b$ but it's still $$ a+ \mathbb{N}\cdot b \subseteq \cup_{i=1}^{n+1}c_i+ \mathbb{N}\cdot d_i$$ It means if you add more $(c,d)$ in the problem then it makes the problem more "weak". I think it will be more accurate when $$ a+ \mathbb{N}\cdot b \subseteq \cap_{i=1}^{n}c_i+ \mathbb{N}\cdot d_i$$