I would do this. Show $$\begin{align}P_t (x,A) &{:=} \int_A g_t (x,y)dy \\
P_0 (x,A) &{:=} \delta_x (A), \end{align}$$
where $g_t$ density of the folded normal distribution, is a Markov kernel (Chapmann K. Equation, etc.).
Then you know there exists a unique Markov process. That's the reflected Brownian motion.
Have someone any comments? Is that enough?
Perhaps you have to show this too.
For all continuous and bounded functions $f$, $$E[f(|B(t+s)|)|F_s]=E[f(|B(t+s)|)|B_s].$$
But this should be $$E[f(|B(t+s)|)|F_s]=E[f(|B(s)+B (s+t)-B (s)|)|F_s]= E[f(|B(t+s)|)|B_s].$$
$$g_{t+s}(x,y)= f_{t+s}(x,y)+ f_{t+s}(x,-y)=\int_R...=\int_{R+} ... +\int_{R-}....=\int_{R+}...= Claim $$
You didn't apply the strong Markov property correctly. This is indicated by the term
$$\mathbb{P}_{X_{\tau}}(|X_{t-\tau}-X_0| \geq r/2)$$
which appears on the right-hand side of $(2)$, note that this term is not well-defined since $t-\tau$ is not a non-negative random variable (it is only non-negative on $\{\tau \leq t\}$), and therefore $X_{t-\tau}$ is not well-defined.
There is the following stronger version of the strong Markov property which is needed here, you can find it for instance in the book by Schilling & Partzsch on Brownian motion (Theorem 6.11)
Theorem Let $(X_t)_{t \geq 0}$ be a $d$-dimensional Brownian motion with canonical filtration $(\mathcal{F}_t)_{t \geq 0}$. Let $\tau$ be a stopping time, and let $\eta \geq \tau$ be an $\mathcal{F}_{\tau+}$-measurable random variable. Then it holds for any bounded Borel measurable function $u$ that $$\mathbb{E}^x (u(X_{\eta}) \mid \mathcal{F}_{\tau+})(\omega) = (T_{\eta(\omega)-\tau(\omega)}u)(X_{\tau}(\omega)) \quad \text{a.s.} \tag{3}$$ where $$(T_s u)(x):= \mathbb{E}^x(u(X_s))$$
The above theorem gives the following corollary:
Corollary Under the assumptions of the theorem it holds that $$\mathbb{E}^x(f(X_{\eta},X_{\tau}) \mid \mathcal{F}_{\tau+})(\omega) = (G_{\eta(\omega)-\tau(\omega)} f)(X_{\tau}(\omega)) \quad \text{a.s.} \tag{4}$$ for any bounded Borel measurable function $f$ where $$(G_s f)(x) := \mathbb{E^x}(f(X_s,x)).$$
The idea is to prove $(4)$ first for functions of the form $f(x,y) = u(x) v(y)$ (... for such functions the assertion is a direct consequence of the above theorem and the pull out property), and then to use a density (or monotone class) argument to extend the identity to any bounded Borel measurable function $f$.
Now for the stopping time $\tau$ from your question define
$$\eta := \max\{t,\tau\}= \begin{cases} \tau, & \{\tau \geq t\}, \\ t, & \{\tau \leq t\}.\end{cases}$$
Then clearly $\eta \geq \tau$ and moreover $\eta$ is $\mathcal{F}_{\tau+}$-measurable. Using the tower property and the pull out property we find that
\begin{align*} \mathbb{P}^x(\tau<t, |X_{t}-X_{\tau}| \geq r/2) &= \mathbb{E}^x \big[ \mathbb{E}^x( 1_{\{\tau <t\}} 1_{\{|X_{t}-X_{\tau}| \geq r/2\}} \mid \mathcal{F}_{\tau+}) \big] \\ &= \mathbb{E}^x \big[ 1_{\{\tau <t\}} \mathbb{E}^x( 1_{\{|X_{t}-X_{\tau}| \geq r/2\}} \mid \mathcal{F}_{\tau+}) \big] \end{align*}
Since $$1_{\{|X_{t}-X_{\tau}| \geq r/2\}} = 1_{\{|X_{\eta}-X_{\tau}| \geq r/2\}} \tag{5}$$ we have $$\mathbb{E}^x( 1_{\{|X_{t}-X_{\tau}| \geq r/2\}} \mid \mathcal{F}_{\tau+}) = \mathbb{E}^x( 1_{\{|X_{\eta}-X_{\tau}| \geq r/2\}} \mid \mathcal{F}_{\tau+}),$$ and therefore it follows from the above corollary that
\begin{align*} \mathbb{P}^x(\tau<t, |X_{t}-X_{\tau}| \geq r/2) &= \int_{\Omega} 1_{\{\tau <t\}}(\omega) \mathbb{P}^{X_{\tau}(\omega)}(|X_{t-\tau(\omega)}-X_0| \geq r/2) \, d\mathbb{P}^x(\omega) \\ &= \int_{(0,t)} \mathbb{E}^x(\mathbb{P}^{X_s}(|X_{t-s}-X_0| \geq r/2); \tau \in ds) \end{align*}
Best Answer
Your question is quite interesting, and deserves more attention than it has been getting. Whether or not a (time homogeneous) Markov process $(X_t)_{t\geq 0}$ with state space $E$ is strong Markov depends on a precise definition. I hope that my explanation below substracts from, rather than adds to, your confusion.
The standard definition of the strong Markov property says that for all times $t\geq 0$, all stopping times $T$, and any bounded, measurable function $f:E\to\mathbb{R}$ we have $$\mathbb{E}(f(X_{t+T}){\bf 1}_{(T<\infty)}\mid{\cal F}_T)=(p_tf)(X_T){\bf 1}_{(T<\infty)},\tag1$$ almost surely, where $p_t$ is the transition function of the process.
In George Lowther's example the transition function is $$p_t(x,\cdot)=N(x,t) \mbox{ for }x\neq0,\quad p_t(0,\cdot)=\delta_0,$$ where $N(x,0)=\delta_x$. The transition function $(p_t)$ satisfies $$p_tf(x)=\cases{f(0)& $x=0$\\[5pt] b_tf(x)& $x\neq 0$}, $$ where $b_t$ is the usual Brownian transition function. This $(p_t)_{t\geq 0}$ is well-defined and time homogeneous.
To be concrete, suppose that $X_0=1$, so that $X_t=W_t+1$ where $W_t$ is a standard Brownian motion. Taking the stopping time $T(\omega)=\inf(t\geq 0: X_t(\omega)=0)$ we have $\mathbb{P}(T<\infty)=1$ and $p_tf(X_T)=p_tf(0)=f(0)$ on the right hand side of (1). To be dramatic, take $f(x)={\bf 1}_{\{0\}}(x)$ so that the right hand side of (1) equals 1.
However, for $t>0$, the left hand side is ${1\over \sqrt{2\pi t}}\int_{-\infty}^\infty f(y+1) \exp(-y^2/2t)\,dy=0$, so the strong Markov property fails. I believe this is the argument that George Lowther had in mind.
What is going on and why does the strong Markov property fail? By changing the transition function at a single point, we have created a disconnect between the process $(X_t)_{t\geq 0}$ and the transition function $(p_t)_{t\geq 0}$. The process is just a shifted Brownian motion that rolls over the point $\{0\}$ and continues to move like a Brownian motion. In contrast, the transition function suggests that the process should get stuck as soon as it hits the point $\{0\}.$
Note that the usual Brownian transition function $(b_t)_{t\geq 0}$ is also a transition function for the process $(X_t)_{t\geq 0}$; transition functions are not unique. With respect to $(b_t)_{t\geq 0}$ our process $(X_t)_{t\geq0}$ does satisfy (1) and therefore is a strong Markov process.
Moral: Whether or not a process is strong Markov, as in (1), depends on which kernel we use. To me, this argues against using the transition function in the definition of strong Markov, and that we ought to use the following transition function-free version instead: $$\mathbb{E}(f(X_{t+T}){\bf 1}_{(T<\infty)}\mid{\cal F}_T)=\mathbb{E}(f(X_{t+T}){\bf 1}_{(T<\infty)}\mid X_T).\tag2$$