The statement of Schur's lema for Lie algebras says that.
Let $(\rho,\mathcal{V})$ is a complex irreducible finite-dimensional representation of a Lie algebra $\mathfrak{g}$. If $T$ conmmutes with $\rho$, then exists $k\in\mathbb{C}$ such that $T = k\operatorname{Id}_{\mathcal{V}}$.
Is the reciprocal satisfied? That is, if the only operators that commute with the representation $\rho$ are multiples of $\operatorname{Id}_{\mathcal{V}}$, then the representation is irreducible?
No; consider the Lie algebra of 2 by 2 upper triangular matrices with the standard Lie bracket and the natural representation on $\mathbb{C}^2$. It is easy to check that this is a counterexample.
However, it is true that $V$ must be indecomposable, i.e., there is no nontrivial decomposition as $V = V_1 \oplus V_2$ where $V_1 \ne V$ and $V_1 \ne 0$. Otherwise the projection $\pi_1: V \rightarrow V_1$ commutes with $\rho$, but is neither an isomorphism nor zero, so cannot be a multiple of the identity.