Part of the difficulty is that $\Delta t$ is not generally well defined. When writing $\Delta E\Delta t$, both terms are meant to be related to variance of operators. However, time is not an operator in quantum mechanics so, as proposed by Mandelstam and discussed in this post, one often defines $$ \Delta t= \frac{\sigma_A}{\vert d\langle A\rangle/ dt\rangle\vert}\, .\tag{1} $$ If there is only one bound state, then $\langle A\rangle $ cannot depend on $t$ because $$ \langle \Psi(t)\vert \hat A\vert \Psi(t)\rangle= \langle \psi\vert\hat A\vert\psi\rangle $$ for $\vert\Psi(t)\rangle=e^{iEt/\hbar}\vert\psi\rangle$.

Thus, for any observable $d\langle A\rangle/dt=0$; in Eq.(1), on can understand $\Delta t\to\infty$ "in the physics way" (assuming $\sigma_A\ne 0$, whatever that means in such a system). You can then qualitatively salvage $\Delta E\Delta t$ by suggesting that $0\times \infty\ge \hbar/2$.

In practice, in a system with only one bound state, there is no natural time scale. You cannot (for instance) define a unit of time based the energy of some transitions because your one bound state cannot transition to any other (bound) state. Thus, the meaning of $\Delta t$ is poorly defined, and one must take great care in interpretation of $\Delta E\Delta t$.

Part of the difficulty is that $\Delta t$ is not generally well defined. When writing $\Delta E\Delta t$, both terms are meant to be related to variance of operators. However, time is not an operator in quantum mechanics so, as proposed by Mandelstam and discussed in this post, one often defines $$ \Delta t= \frac{\sigma_A}{\vert d\langle A\rangle/ dt\rangle\vert}\, .\tag{1} $$ If there is only one bound state, then $\langle A\rangle $ cannot depend on $t$ because $$ \langle \Psi(t)\vert \hat A\vert \Psi(t)\rangle= \langle \psi\vert\hat A\vert\psi\rangle $$ for $\vert\Psi(t)\rangle=e^{iEt/\hbar}\vert\psi\rangle$.

Thus, for any observable $d\langle A\rangle/dt=0$; in Eq.(1), one can understand $\Delta t\to\infty$ "in the physics way" (assuming $\sigma_A\ne 0$, whatever that means in such a system). You can then qualitatively salvage $\Delta E\Delta t$ by suggesting that $0\times \infty\ge \hbar/2$.

In practice, in a system with only one bound state, there is no natural time scale. You cannot (for instance) define a unit of time based the energy of some transitions because your one bound state cannot transition to any other (bound) state. Thus, the meaning of $\Delta t$ is poorly defined, and one must take great care in interpretation of $\Delta E\Delta t$.