Do there exist elementarily equivalent models $\mathcal{M}, \mathcal{N}$ of the cardinality $\kappa=\omega$ such that neither can be elementarily embedded into the other?
If the models were not required to be of the cardinality $\omega$, then one could consider models of the theory of 3 equivalence classes where each class has to be infinite: $\mathcal{M}$ would have its domain $M = M_1 \sqcup M_2 \sqcup M_3$ and $dom(\mathcal{N})=N_1 \sqcup N_2 \sqcup N_3$ where $M_1, M_2, M_3, N_1, N_2, N_3$ are equivalence classes of their respective cardinalities $\omega_1, \omega_1, \omega_2, \omega_0, \omega_2, \omega_2$. Because of the incompatible cardinalities, there is no embedding from $\mathcal{M}$ to $\mathcal{N}$ or vice versa. Therefore this is the answer to my question for the cardinality $\kappa \gt \omega_1$. For $\kappa=\omega_1$ one could use a similar trick with equivalence classes over two relations and the incompatibility of the cardinalities $\omega_0, \omega_1$. But what about the cardinality $\kappa=\omega_0$?
The language contains a relation symbol $<$ and the constants $a_i,b_i$ for $i\in\omega$. Consider two models, both with domain $\mathbb Q$ and the natural interpretation of $<$. In both models $b_{i+1}<b_i<a_i<a_{i+1}$. In the first model $a_i\to+\infty$ and $b_i\to-1$ in the second model $a_i\to+1$ and $b_i\to-\infty$.
By quantifier elimination the two models are elementary equivalent. Clearly, they do not embed into each other.