 Solution

Assume first that $S\ne 0$, as required for the irreducible-module conclusion. If $B\to {\mathrm{End}}_{\mathbb{C}}(S)$ is surjective, any nonzero invariant subspace contains every image of one of its nonzero vectors under all endomorphisms, hence is all of $S$. Thus $S$ is irreducible.

Conversely an irreducible $B$-module is finite dimensional: for $s\ne 0$, $S=Bs$ is a quotient of the finite-dimensional vector space $B$. Let $D$ be the image of $B$ in ${\mathrm{End}}_{\mathbb{C}}(S)$. It acts faithfully and irreducibly. The subspace $J(D)S$ is a submodule, so is zero or all of $S$. The latter alternative would imply $J(D{)}^{k}S=S$ for every $k$, contradicting nilpotence of the Jacobson radical. Therefore $J(D)S=0$, and faithfulness gives $J(D)=0$.

The Artin–Wedderburn theorem makes $D$ a product of matrix algebras. A faithful simple module forces there to be just one factor, because all other factors would annihilate that module. Thus $D\cong M_d(\mathbb{C})$ and $S\cong {\mathbb{C}}^d$, so the action is the full endomorphism algebra. This is the Burnside matrix-algebra theorem.

For nonzero $S$, surjectivity is equivalent to irreducibility. The zero module is a literal exception if it is admitted: its endomorphism algebra is zero, so the action map is surjective, whereas the zero module is not irreducible.