We define a sequence of (discrete) stopping times
$$\tau_j := \frac{\lfloor 2^j \tau \rfloor+1}{2^j}, \qquad j \in \mathbb{N}.$$
It is not difficult to see that $\tau_j$ is indeed a stopping time and $\tau_j \downarrow \tau$ as $j \to \infty$. Since the Brownian motion has continuous paths, this implies $B(\tau) = \lim_{j \to \infty} B(\tau_j)$.
Let $\xi,\eta \in \mathbb{R}$. Then, by the dominated convergence theorem,
$$\begin{align*} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg) &= \lim_{j \to \infty} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau_j+t)-B(\tau_j))} \cdot e^{\imath \, \eta B(\tau_j)} \bigg) \\ &= \lim_{j \to \infty} \sum_{k=1}^{\infty} \mathbb{E} \bigg( e^{\imath \, \xi (B(k \cdot 2^{-j} +t)-B(k \cdot 2^{-j}))} \cdot e^{\imath \, \eta B(k \cdot 2^{-j})} \cdot 1_{\{\tau_j = k \cdot 2^{-j}\}} \bigg) \end{align*}$$
where we used in the last step that $\tau_j$ is a discrete stopping time. By assumption, $B(k \cdot 2^{-j}+t)-B(k \cdot 2^{-j})$ and $B(k \cdot 2^{-j}) \cdot 1_{\{\tau_j=k2^{-j}\}}$ are independent. Therefore, we obtain
$$\begin{align*} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg) &= \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg) \lim_{j \to \infty} \sum_{k=1}^{\infty} \mathbb{E} \bigg( e^{\imath \, \eta B(k 2^{-j})} \cdot 1_{\{\tau_j=k \cdot 2^{-j}\}} \bigg) \\ &= \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg) \cdot \mathbb{E}\bigg(e^{\imath \, \eta B(\tau)} \bigg). \end{align*}$$
(In the second step we used again dominated convergence, similar to the above calculation.) If we choose $\eta = 0$, then we get
$$ \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \bigg) = \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg);$$
hence,
$$ \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg)= \mathbb{E}\bigg(e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) \cdot \mathbb{E}\bigg(e^{\imath \, \eta B(\tau)} \bigg)$$
i.e. $B(\tau+t)-B(\tau)$ and $B(\tau)$ are independent. Therefore, the strong Markov property gives
$$\begin{align*} \mathbb{E}\bigg(1_F e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) &= \mathbb{E}\bigg(1_F \mathbb{E} \bigg[ e^{\imath \, \xi (B(\tau+t)-B(\tau))} \mid \mathcal{F}_{\tau} \bigg] \bigg) \\ &= \mathbb{E}\bigg(1_F \mathbb{E} \bigg[ e^{\imath \, \xi (B(\tau+t)-B(\tau))} \mid B_{\tau} \bigg] \bigg)\\ &= \mathbb{P}(F) \cdot \mathbb{E}\bigg(e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) \end{align*}$$
for any $F \in \mathcal{F}_{\tau}$. Consequently, $B(\tau+t)-B(\tau)$ is independent of $\mathcal{F}_{\tau}$.
A very similar calculation shows that
$$\mathbb{E} \left( \exp \left( \imath \sum_{j=1}^n \xi_j \cdot (B(\tau+t_j)-B(\tau+t_{j-1})) \right) \right) = \prod_{j=1}^n \mathbb{E}e^{\imath \, \xi_j B(t_j-t_{j-1})}$$
for any $\xi_j \in \mathbb{R}$, $0 \leq t_0 < \ldots \leq t_n$. This means that $(B(\tau+t_j)-B(\tau+t_{j-1}))_{j=1,\ldots,n}$ are independent normal distributed random variables.
First of all, note that the natural filtration of $(X_t)_{t \geq 0}$ equals the natural filtration of $(W_t)_{t \geq 0}$; this follows directly from the relation
$$X_t = \exp \left( \left[ \mu - \frac{\sigma^2}{2} \right] t + \sigma W_t \right).$$
The Markov property of $(X_t)_{t \geq 0}$ can be proved as follows: Fix some bounded Borel-measurable function $f$ and $s \leq t$. For brevity, set $c := \mu - \frac{\sigma^2}{2}$. Then
$$\begin{align*} \mathbb{E}(f(X_t) \mid \mathcal{F}_s) &= \mathbb{E} \bigg( f \left[ e^{ct} e^{\sigma(W_t-W_s)} e^{\sigma W_s} \right] \mid \mathcal{F}_s \bigg). \end{align*}$$
Since $W_t-W_s$ is independent from $\mathcal{F}_s = \sigma(W_r; r \leq s)$ and $W_s$ is $\mathcal{F}_s$-measurable, we get
$$\mathbb{E}(f(X_t) \mid \mathcal{F}_s) = g(e^{ct} e^{\sigma W_s}) \tag{1}$$
where
$$g(y) := \mathbb{E} \left( f(y e^{\sigma (W_t-W_s)}) \right).$$
Since the right-hand side of $(1)$ is $X_s$-measurable, the tower property yields
$$\mathbb{E}(f(X_t) \mid X_s) = \mathbb{E} \bigg[ \mathbb{E}(f(X_t) \mid \mathcal{F}_s) \mid X_s \bigg] \stackrel{(1)}{=} g(e^{ct} e^{\sigma W_s}). \tag{2}$$
Combining $(1)$ and $(2)$ we find
$$\mathbb{E}(f(X_t) \mid X_s) = \mathbb{E}(f(X_t) \mid \mathcal{F}_s).$$
Best Answer
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)
The above theorem gives the following corollary:
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*}