I'll discuss this in the context of Yukawa theory, and use renormalized perturbation theory. The way I understand it is this.
Setup: Consider the renormalization scheme
$$ \psi_R = \frac{1}{\sqrt{Z}}\psi_0 \\ m_R = \frac{1}{Z_m} m_0 $$
where $Z_i = 1 + \delta_i$. Then the renormalized propagator is.
$$G^R = \frac{1}{Z_2} G^{bare}. $$
Now recall that summing over all 1PI insertions we may write
$$iG^{bare} = \frac{i}{\not{p} - m_0 + \Sigma(\not{p})}$$
where $\Sigma$ is the sum over all 1PI insertions. Now, we can write $\Sigma = \Sigma_2 + {\cal O}(g^4)$ keeping only those insertions of order less than $g^4$. So that our renormalized propagator to order $g^2$ is
$$G^R = \frac{1}{1+\delta}\frac{i}{\not{p} - m_0 + \Sigma_2(\not{p})} \\ \boxed{G^R= \frac{i}{\not{p} - m_R + \Sigma_R(\not{p})} } $$
where we have defined
$$ \Sigma_R = \Sigma_2 + \delta \not{p} - (\delta + \delta_m)m_R $$
Now, recall that we define the physical mass as the pole of the propagator. Therefore,
$$ m_P - m_R + \Sigma_R(m_P) = 0 .$$
This implies that
$$ \boxed{\delta_m m_P = \Sigma_R(m_P)} $$
Your question: Let us call $\Sigma_2(\not{p}) = i\Delta m$, anticipating that the correction is purely imaginary. Then, our propagator is
$$\tilde{G}^R= \frac{i}{\not{p} - m_P + i\Delta m} $$$$\tilde{G}^R= \frac{i}{\not{p} - m_R+ i\Delta m} $$
where now we know that our propagator has a pole at $m_P$.
Recall that the amplitude for a particle to propagate from $x$ to $y$ is given by the fourier transform
$$D(x-y) = \int \frac{d^4p}{(2\pi)^4} \frac{i}{\not{p} - m_P + i\Delta m}e^{-ip(x-y)} $$$$D(x-y) = \int \frac{d^4p}{(2\pi)^4} \frac{i}{\not{p} - m_R + i\Delta m}e^{-ip(x-y)} $$
To make things easier suppose that y=0. Then we have
$$D(x) = \int \frac{d^4p}{(2\pi)^4} \frac{i}{\not{p} - m_P + i\Delta m}e^{i\mathbf{p\cdot x}}e^{-iEt} $$$$D(x) = \int \frac{d^4p}{(2\pi)^4} \frac{i}{\not{p} - m_R+ i\Delta m}e^{i\mathbf{p\cdot x}}e^{-iEt} $$
Where $E = \sqrt{p^2 + m_{rest}^2}$. What is the mass of the particle in its rest frame? Well by definition it's the pole of the propagator! And so we may replace
$$E = \sqrt{p^2 + (m_R - i \Delta m)^2} \\ =\sqrt[4]{\left(-{\Delta m}^2+m^2+p^2\right)^2+4 {\Delta m}^2 m^2} \left(i \sin \left(\frac{\phi}{2}\right)+\cos \left(\frac{\phi}{2}\right)\right)\\ \boxed{E \equiv \xi \left(i \sin \left(\frac{\phi}{2}\right)+\cos \left(\frac{\phi}{2}\right)\right) }$$
where $\phi$ is the $\mathrm{Arg}$ of the radicand.Finally, we obtain that
$$ \boxed{D(x) = \int \frac{d^4p}{(2\pi)^4} \frac{ie^{i\mathbf{p\cdot x}} e^{-i\xi\cos(\frac{\phi}{2}) t}}{\not{p} - m_P + i\Delta m}e^{-\xi\sin(\frac{\phi}{2}) t}} $$$$ \boxed{D(x) = \int \frac{d^4p}{(2\pi)^4} \frac{ie^{i\mathbf{p\cdot x}} e^{-i\xi\cos(\frac{\phi}{2}) t}}{\not{p} - m_R+ i\Delta m}e^{-\xi\sin(\frac{\phi}{2}) t}} $$
and so we see that indeed the probability for a particle to exist at a time $t$ decays exponentially.