People usually say $V \subset H = H^* \subset V^*$ is a Gelfand triple if $V$ is continuously and densely embedded in $H$ and $H$ is identified with its dual.
Sometimes they do not mentioned that $H$ is identified with its dual. See Zeidler, section 23.4, where every $h \in H$ is identified with $\bar h \in V^*$ where $\langle \bar h, v \rangle_{V^*,V} := (h, v)_H$.
What goes wrong if we do not identify $H$ with its dual? Of course I understand that $H^* \subset V^*$.
The isomorphism $H \to H^*$ is always there, whether you pay attention to it or not. Making it explicit would be annoying; it would be akin to disallowing yourself to treat a rational number as if it were a real number, and having to add extra notation whenever you want to convert a rational number to its corresponding real number.