It is known from the Strong Markov property of Brownian motion that, given a Brownian motion $B$ and the a.s. finite stopping time $\tau_b := \inf \{ t \ge 0: B_t = b \}$, $$ (B_{\tau_b + t} - B_{\tau_b})_{t \ge 0} $$ is independent of the stopped sigma algebra $$ \mathcal{F}_{\tau_b} := \left \{A \in \sigma \left (\cup \mathcal F_t \right )\mid \forall t: \{ \tau_b \le t \} \cap A \in \mathcal F_t \right \}. $$
Does it also hold that $$ (B_{\tau_b + (t - \tau_b)} - B_{\tau_b})_{t \ge 0} = (B_t - B_{\tau_b})_{t \ge 0} $$ is independent of $$ \mathcal{F}_{\tau_b} \quad ? $$
Addendum
Looking at the second edition of the book this question came from (Brownian motion by R. Schilling and L. Partzsch), instead of the first edition, I found that the equation that confused me have been changed. Now the claim seems rather to be that the probabilities of the sets
$$
\left\{\tau_b \le t \right \} \bigcap \left\{ B_{\tau_b + (t - \tau_b)} - B_{\tau_b} < 0 \right\}
$$
and
$$
\left\{\tau_b \le t \right \} \bigcap \left\{ B_{\tau_b + (t - \tau_b)} - B_{\tau_b} > 0 \right\}.
$$
are equal. Though some reference still seems to be made to that
$$
\left\{ B_{\tau_b + (t - \tau_b)} - B_{\tau_b} < 0 \right\} \in \mathcal{F}_\infty^W := \sigma\left (\bigcup B_{\tau_b + t} - B_{\tau_b} \right )
$$
(but perhaps it isn't claimed that the whole process $B_{\tau_b + t} - B_{\tau_b}$ is measurable w.r.t. $\mathcal F_\infty^W$ and thus independent of $\mathcal F_{\tau_b}$).
Here is an excerpt from (the second edition of) the book
I would still like help to verify that $$ \mathrm P \left\{\tau_b \le t \right \} \bigcap \left\{ B_{\tau_b + (t - \tau_b)} - B_{\tau_b} < 0 \right\} = \mathrm P \left\{\tau_b \le t \right \} \bigcap \left\{ B_{\tau_b + (t - \tau_b)} - B_{\tau_b} > 0 \right\}. $$
In your original formulation, I believe the result is false, since $B_{t}-B_{\tau_b} = B_{t} -b$ cannot be equal to $0$ prior to time $\tau_b$, while it must tend to $0$ as $t\to\tau_b$.
In the revised version of the book, the statement seems more clear. Note that we restrict ourselves to the event that $\tau_b \leq t$, so we then know that $B_{t}-B_{\tau_b}$ is independent of $\mathcal{F}_{\tau_b}$ by the strong Markov property. I.e., we are now looking into the past, as opposed to the ambiguous case in the first version. Also, by the strong Markov property we have that $(B_{t}-B_{\tau_b})_{t\geq \tau_b}$ is a Brownian motion, implying that $B_{t}-B_{\tau_b}\sim N(0,t - \tau_b)$. So then by symmetry of the normal distribution we have $$ \mathbb{P}\left(\lbrace \tau_b\leq t\rbrace\cap\lbrace B_{t}-B_{\tau_b} < 0 \rbrace\right)= \mathbb{P}\left( \lbrace \tau_b \leq t\rbrace\cap\lbrace B_{t}-B_{\tau_b} > 0\rbrace \right) $$ Now just substitute $B_{t}=B_{\tau_b + (t-\tau_b)}$ and I believe we are done. The crucial step seems to be that restrict ourselves to $\lbrace \tau_b \leq t\rbrace$.
Edit: Regarding the symmetry of the normal distribution, it is of course not true in general, that $\mathbb{P}(A\cap \lbrace B_{t} <0 \rbrace) = \mathbb{P}(A\cap \lbrace B_{t} > 0\rbrace)$ for any $A$. The following argument is used in Le Gall's book on Brownian motion and stochastic calculus:
We have $(\tau_b , B_{\tau_b + t - \tau_b} - B_{\tau_b}) \overset{d}{=}(\tau_b, B_{\tau_b} - B_{\tau_b + t - \tau_b})$ where this distribution is the product of the law of $\tau_b$ with the Wiener measure on $C(\mathbb{R}_{+},\mathbb{R})$. (technically I guess we need an indicator function to ensure finiteness of the stopping time). Then set $$ H = \lbrace (s,w)\in \mathbb{R}_{+}\times C(\mathbb{R}_{+},\mathbb{R})\mid s\leq t, w(t-s) \leq 0 \rbrace $$ where $w\mapsto w(t)$ is the coordinate map on the space of continuous functions on the positive half-line. Then we have $$ \mathbb{P}((\tau_b,B_{\tau_b + t - \tau_b}-B_{\tau_b})\in H) = \mathbb{P}((\tau_b,-(B_{\tau_b + t - \tau_b}-B_{\tau_b}))\in H) $$ and also $$ \mathbb{P}((\tau_b,B_{\tau_b + t - \tau_b}-B_{\tau_b})\in H) = \mathbb{P}(\tau_b \leq t, B_{\tau_b + t - \tau_b}- B_{\tau_b}\leq 0) $$ from where the desired symmetry follows.