Let $K$ be a ring and let $L$ be the extension ring of $K$ given by $L=K(a_1,\ldots,a_n)$
Question 1: Is there a ring $L$ such that it does not have any prime element belong to $\mathbb{Z}$ $(2,3,5,7,11,13,\ldots\notin P_L $ where $ P_L $ is the set of the prime element of $L$).
Question 2 : is there a ring $L$ in which if $p \in P_{L}$ then $ p\nmid n $ for any integer $ n \in \mathbb{N} $ (or a lest some particular residue class like $n \equiv a \bmod d $ for some $a, d \in \mathbb{N} $ ?
Question 3: Is there a ring which satisfies question 1 and 2 at the same time?
To elaborate on the comment of @reuns. I will assume that $\mathbb{Z}$ is a subring of your '$K$' (although I am going to change this notation) otherwise the question doesn't make sense. If I interperate your first question correctly then the answer is 'yes'.
Consider the ring $A = \mathbb{Z}$ and the ring $B = \mathcal{O}_K$ where $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$. Explicitly we may write $B = \mathbb{Z}[\sqrt{2}, \sqrt{3}]$.
My argument requires a bit of number theory - perhaps someone can find an elementary argument. When $p = 2$ or $3$ it is easy to see that $2B = (\sqrt{2})^2$ and similarly for $3$, so we may assume that $p$ is unramified in $K$.
Take some prime ideal $\mathfrak{p}$ of $B$ lying above $p$. We know that $p$ is inert (i.e., $pB$ is prime) if and only if the decomposition group $D_{\mathfrak{p}/p} \subset \operatorname{Gal}(K/\mathbb{Q}) \cong (\mathbb{Z}/2\mathbb{Z})^2$ is the entire group . But $p$ is unramified, so $D_{\mathfrak{p}/p}$ is isomorphic to the galois group of the extension of reside fields, which of course is cyclic (since the residue fields are finite). From this we see that $pB$ cannot be a prime ideal for any prime $p$ of $\mathbb{Z}$.
The answer to the second is 'No'. If $\pi$ is a prime element of a ring $B \supset \mathbb{Z}$ then $(\pi) \cap \mathbb{Z}$ is a prime ideal. In particular $p \in (\pi)$ for some prime $p$.
A few points on your post: $K$ and $L$ are normally reserved for fields we tend to use $A$, $B$, $R$ etc., for rings. If $A$ is a ring and you write $A(x)$ you are implicitly adjoining inverses too.