Yes there are other indecomposable modules where $L_0$ is bounded from below. Logarithmic theories provide many examples.

The simplest example I can think of: take a Verma module generated by a primary state of conformal dimension $\Delta$, and formally take the derivative of this state with respect to $\Delta$. The derived state generates a logarithmic module (i.e. $L_0$ is not diagonalizable) where $L_0$ is still bounded from below.

This example works for any $\Delta$. For special values of $\Delta$ such that Verma modules have singular vectors, more complicated constructions are also possible, including non-logarithmic examples. For example, consider a primary state $v_0$ of dimension zero, and a state $v_1$ such that
$$
L_{n\geq 2} v_1 = 0 \quad , \quad L_1 v_1 = v_0 \quad , \quad L_0 v_1 = v_1
$$
Then the representation generated by $v_1$ is not a Verma module, as it is strictly larger than the Verma module generated by $v_0$. Since $v_1$ is an $L_0$ eigenvector, $L_0$ is diagonalizable in that representation. And our representation is indecomposable thanks to our choice of dimension for $v_0$, which prevents us from decomposing it using a subrepresentation generated by a linear combination of $v_1$ and $L_{-1}v_0$.