The answer is no. Take a non-regular language $L$ on the alphabet $A = \{a, b\}$, say $L = \{a^nb^n \mid n \geqslant 0\}$ and consider the family ${\cal L} = \{L^*\}$. Then $\cal L$ is closed under union, product and star, but it is not closed under intersection with regular sets (take the intersection with any cofinite language), homomorphisms (take the morphism $\varphi$ defined by $\varphi(u) = 1$ for every word $u$), inverse of homomorphisms (take the same $\varphi$ and observe that $\varphi^{-1}(L^*) = A^*$).
Let me also answer your first question "Why can regular expressions be defined without mentioning closure ...". Regular languages are sometimes defined using regular expressions and sometimes using automata. There is no harm as long as these two definitions define the same class. This is Kleene's theorem, which implies that regular languages are closed under intersection and complement. However, these two definitions are no longer equivalent in the more general setting described below.
Let $M$ be a monoid. The rational subsets of $M$ are those defined from singletons using the operations of finite union, product and star. Here the product of two subsets $S$ and $T$ is the set $ST = \{st \mid s \in S, t \in T\}$ and $S^*$ is the submonoid of $M$ generated by $S$. But now, rational sets are not necessarily closed under intersection or complement. They are closed under monoid homomorphisms, but not necessarily under inverses of homomorphisms.
A subset $P$ of $M$ is recognizable if there is a finite monoid $F$ and a monoid homomorphism $f: M \to F$ such that $P = f^{-1}(f(P))$. One can show that recognizable sets are closed under finite union, finite intersection, complement and inverse of homomorphisms. However, there are not necessarily closed under homomorphisms.
To come back to languages, if $M$ is a free monoid, the two notions, rational and recognizable, coincide and both define the regular sets.
