It is perhaps worth pointing out the special case that if the semidirect product $GH$ is a Frobenius group with kernel $G$ and complement $H$, then $(GH,H)$ is a Gelfand pair if and only if $G$ is Abelian.
In general, we have a Gelfand pair from such a direct product if and only if the permutation character $\theta = {\rm Ind}_{H}^{GH}(1)$ is multiplicity free. One explanation why we always get a Gelfand pair when $G$ is Abelian is that even the restriction ${\rm Res}^{GH}_{G}(\theta)$ is the regular character of $G$, which is multiplicity free when $G$ is Abelian.
Suppose now that $GH$ is such a Frobenius group, but that $G$ is non-Abelian. Then $G$ has a non-linear irreducible character $\mu$, which occurs with multiplicity $\mu(1)$ in ${\rm Res}^{GH}_{G}(\theta)$. On the the other hand, there is just one irreducible character $\chi$ of $G$ which contains $\mu$ with non-zero multiplicity on restriction to $G$, and that is $\chi = {\rm Ind}^{GH}_{G}(\mu)$. Hence $\chi$ must occur with multiplicity $\mu(1) > 1$ in $\theta $, so that $(GH,H)$ is not a Gelfand pair.
More generally, when $H$ is cyclic, or when $H$ is Abelian of order relatively prime to $|G|$, it is possible to use Clifford's theorem to check that $(GH,H)$ is not a Gelfand pair if there is an irreducible character $\mu$ of $G$ such that $\mu(1) > |I_{H}(\mu)|$, where $I_{H}(\mu) = \{h \in H : \mu^{h} = \mu \}$, the Frobenius group case being an extreme example of this where $I_{H}(\mu)$ is the trivial group.