Convergence to $0$ is simple:
Proposition. The following are equivalent:
(i) The operator $e^{tQ}$ converges to $0$ with respect to the operator norm as $t \to \infty$.
(ii) The spectrum of $Q$ is contained in the open left halfplane $\{\lambda \in \mathbb{C}: \, \operatorname{Re} \lambda < 0\}$.
Sketch of proof. As mentioned by Pietro Majer in the comments, this follows from the spectral mapping theorem for the operator exponential function, along with the spectral radius formula. $\square$
Convergence to a non-zero operator is a bit more involved. Here are the details:
Theorem. The following are equivalent:
(i) The operator $e^{tQ}$ converges with respect to the operator norm to a non-zero operator as $t \to \infty$.
(ii) The spectrum of $Q$ is contained in the set $\{\lambda \in \mathbb{C}: \, \operatorname{Re} \lambda < 0\} \cup \{0\}$, and $0$ is an isolated point in the spectrum and a first order pole of the resolvent of $Q$.
Sketch of proof. This also follows from classical spectral theory, for instance by splitting the space into the kernel and the range of the mean ergodic projection of the semigroup $(e^{tQ})_{t \ge 0}$. (I can add a few more details here if required.) $\square$