Can the definition of regular languages be simplified?
No. Closure under concatenation means that, if $A$ and $B$ are regular languages, then so is the language of strings formed by concatenating taking one string from $A$ and one from $B$. Closure under Kleene star means that, if $A$ is a regular language, then so is the langauge formed by taking any finite number of strings from $A$ and concatenating them.

If you delete Kleene star then every regular language would be finite. The reason is that any given regular language must be formed by a finite number of operations from the basic languages ($\emptyset$ and singletons). But any language whose definition involves no Kleene stars and only $k$ concatenations can only contain strings of length at most $k$, so it must be finite (it has at most $|\Sigma|^k$ strings in it).

Since there are infinite regular languages, the languages formed without Kleene star are a strict subset of the regular languages.