$$\mathbb{P}(B^x_t\in A)=\mathbb{P}(B^x_t\in A,B^x_s>0,\forall s\in [0,t])+\mathbb{P}(B^x_\tau=0,B^x_t\in A)$$
where $$\tau=inf\{s,B^x_s=0,s\in[0,t]\}$$
And
$$\mathbb{P}(B^x_\tau=0,B^x_t\in A)=\mathbb{P}(B^x_\tau=0,B^x_t\in -A)=\mathbb{P}(B^x_t\in -A)$$
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:
$$\mathbb{E}(f(X_{t+T}){\bf 1}_{(T<\infty)}\mid{\cal F}_T)=\mathbb{E}(f(X_{t+T}){\bf 1}_{(T<\infty)}\mid X_T).\tag2$$
Best Answer
Let's consider separately the case $x=0$ and $x \neq 0$.
If $x=0$, then $X_t=0$ for all $t \geq 0$, and so
$$\mathbb{E}(1_B(X_{t+s}) \mid \mathcal{F}_s) = \mathbb{E}(1_B(0) \mid \mathcal{F}_s) = 1_B(0)$$ for all $s,t \geq 0$. Exactly the same reasoning gives
$$\mathbb{E}(1_B(X_{t+s}) \mid X_s) = \mathbb{E}(1_B(0) \mid X_s) = 1_B(0).$$ Hence, $$\mathbb{E}(1_B(X_{t+s}) \mid \mathcal{F}_s) = \mathbb{E}(1_B(X_{t+s}) \mid X_s).$$
If $x \neq 0$, then $X_t=x+W_t$ for some Brownian motion $(W_t)_{t \geq 0}$ started at $0$. Note that the canonical filtration of $(X_t)_{t \geq 0}$ is the canonical filtration of $(W_t)_{t \geq 0}$, i.e. there is no need to distinguish between the two filtrations. By the Markov property of Brownian motion, we have
\begin{align*} \mathbb{E}(1_B(X_{t+s}) \mid \mathcal{F}_s) &= \mathbb{E}(1_B(x+W_{t+s}) \mid \mathcal{F}_s) = g(W_s) \end{align*} for $$g(y) := \mathbb{E}(1_B(x+y+W_t)).$$
Equivalently,
$$\mathbb{E}(1_B(X_{t+s}) \mid \mathcal{F}_s) = g(W_s+x-x) = g(X_s-x).\tag{1}$$
Since the right-hand side is $\sigma(X_s)$-measurable, it follows from the tower property of conditional expectation that
\begin{align*}\mathbb{E}(1_B(X_{t+s}) \mid X_s) &= \mathbb{E} \big[ \mathbb{E}(1_B(X_{t+s}) \mid \mathcal{F}_s) \mid X_s \big] \\ &\stackrel{(1)}{=} \mathbb{E}(g(X_s-x) \mid X_s) = g(X_s-x). \tag{2} \end{align*}
Hence,
$$\mathbb{E}(1_B(X_{t+s}) \mid \mathcal{F}_s) \stackrel{(1)}{=} g(X_s-x) \stackrel{(2)}{=} \mathbb{E}(1_B(X_{t+s}) \mid X_s). $$