This is again a question about this example (also see here).
It seems we can write this process as $$X_t := \bigl(t - \text{Exp}(1) \mathbf{1}_{\{X_0 = 0\}}\bigr)^+ + \mathbf{1}_{\{X_0 \neq 0 \}}X_0\, .$$
Where $\text{Exp}(\lambda)$ is the exponential distribution (CDF: $1-e^{-\lambda x}$) with parameter $\lambda$.
Define the stopping time $\tau := \inf \{ t\geq 0 : X_t > 0\}$.
I've read the following explanation why this is not strong Markov:
$$\mathbf{E}_0 \bigl[f((X_{\tau + t}))|\mathcal{F}_\tau\bigr] = f(t) \neq \int f(y) \kappa_t(0, dy) = e^{-t} f(0) + \int_0^t e^{-s} f(t-s)\, ds\, ,$$
but that makes no sense to me... how can it be $f(t)$? Because $t \in I$ doesn't fit into a function $f\colon E^{\otimes I}\rightarrow \mathbb{R}$ (see here for the definition of strong Markov property. For this example $E = [0, \infty)$)...