No. For a counterexample let $H=(\omega,E)$ where $E=\{e_n:n\in\omega\}$ and $e_n=[n,\omega)=\{x\in\omega:x\ge n\}$.
If a subset $E'\subseteq E$ is finite then $(\omega,E')$ is tameable; every vertex cover contains a finite vertex cover which contains a minimal vertex cover, and every independent set is contained in a cofinite independent set which is contained in a maximal independent set.
If $E'\subseteq E$ is infinite then $(\omega,E')$ is not tameable; in fact, it has no minimal vertex cover and no maximal independent set, since the vertex covers are just the infinite subsets of $E$ and the independent sets are just the coinfinite subsets of $E$.
Thus $\operatorname{Tame}(H)=\{E'\subseteq E:E'\text{ is finite}\}$ and $\operatorname{Max}(\operatorname{Tame}(H))=\varnothing$.