If $L$ has characteristic polynomial $\lambda \mapsto p(\lambda)$, then the conjugated differential operator $e^{-qt} L e^{qt}$ has characteristic polynomial $\lambda \mapsto p(\lambda+q)$. From this we may easily reduce the problem to the $q=0$ case.
If the characteristic polynomial $\lambda \mapsto p(\lambda)$ of $L$ has a zero of order $k$ at the origin, then $p$ factors as $p(\lambda) = \tilde p(\lambda) \lambda^k$, and $L$ similarly factors as $L = \tilde L \frac{d}{dt^k}$. Since every degree $m$ polynomial has a $k$-fold antiderivative that is equal to $t^k$ times a degree $m$ polynomial, we can thus reduce the problem to the $q=0, m=0$$q=0, k=0$ case.
If $q=0$ and $m=0$$k=0$, then $p(0)$ is non-vanishing; by rescaling we may take $p(0)=1$. Then $p(\lambda) = 1 + \lambda r(\lambda)$ for some polynomial $r$, so $L = 1 + R \frac{d}{dt}$ for some differential operator $R$. On the finite dimensional vector space spanned by $1,t,\ldots,t^m$, the operator $R \frac{d}{dt}$ acts nilpotently (as it always reduces the degree) and so $L$ is unipotent, hence invertible (by Neumann series), in this space, and the claim follows.