Is Rice-Shapiro theorem bidirectional?
No, the converse direction does not hold:

Pick some non-c.e. set $B \subseteq \mathbb{N}$. Now consider $\Gamma = \{A \ c.e. \mid A \cap B \neq \emptyset\}$. By construction, $\Gamma$ satisfies the conclusion of Rice-Shapiro (it suffices to consider singleton sets rather than all finite sets). However, for the resulting index set $I$ we find that $B \leq_m I$, so $I$ cannot be c.e..

Essentially, Rice-Shapiro says that the Markov-effectively open subsets of the subspace of ce subsets of $P(\omega)$ (with the Scott topology) are open, but for already for cardinality reasons alone, not all open subsets can be Markov-effectively open.