$$\mathbb{P}(B^x_t\in A)=\mathbb{P}(B^x_t\in A,B^x_s>0,\forall s\in [0,t])+\mathbb{P}(B^x_\tau=0,B^x_t\in A)$$
where $$\tau=inf\{s,B^x_s=0,s\in[0,t]\}$$
And
$$\mathbb{P}(B^x_\tau=0,B^x_t\in A)=\mathbb{P}(B^x_\tau=0,B^x_t\in -A)=\mathbb{P}(B^x_t\in -A)$$
The generator of a sum of (time- and space-) homogeneous independent Markov processes is a sum of generators.
Indeed, let $X$ and $Y$ be these processes. Denote by $T_t$ and $S_t$ their Markov semigroups, $A$ and $B$ their generators, $\mathsf E_{x,y}$ the expectation given $X_0=x, Y_0=y$. Then for any $f\in C_b(\mathbb R)$ (bounded continuous), using the independence and homogeneity,
$$
\mathsf E_{x,y}[f(X_t + Y_t)] = \mathsf E_{x,y}[\mathsf E_{x,y}[f(X_t + z)]|_{z=Y_t}]= \mathsf E_{x,y}[\mathsf E_{x+z,y}[f(X_t)]|_{z=Y_t}]
\\ = \mathsf E_y[T_t f(x+z)|_{z=Y_t}] = S_t T_tf (x+y).
$$
Similarly, it equals to $T_t S_t f (x+y)$, in particular, $T_t$ and $S_t$ commute (and so do $A$ and $B$). Now for $f\in C_b^2(\mathbb R)$ (bounded twice continuously differentiable with bounded derivatives), differentiating this equality in $t$ for $t=0$, we get
$$
\frac{d}{dt}\mathsf E_{x,y}[f(X_t + Y_t)]|_{t=0} = (A+B)f(x+y),
$$
which means that $A+B$ is the generator of $X+Y$.
There is another way to look at this. A time- and space-homogeneous process is a Lévy process, which can be written as a sum
$$
at + b W_t + Z_t,
$$
where $W_t$ is a standard Wiener process, and $Z_t$ is a pure jump process. When you add two independent things of such kind, you arrive to
$$
(a_1+a_2)t + \sqrt{b_1^2 + b_2^2}\, W_t + (Z_t^1 + Z_t^2).
$$
The process $Z_t = Z_t^1 + Z_t^2$ is again a pure jump process, and the jump intensities at a given level just add up. If you translate these things to generator, everything will add up to; for instance, the diffusion part is
$\left(\sqrt{b_1^2 + b_2^2}\right)^2 \frac{d^2}{dx^2} = (b_1^2 + b_2^2)\frac{d^2}{dx^2}$.
Best Answer
Denote by $$p_t(y) := \frac{1}{\sqrt{2\pi t}} \exp \left(-\frac{y^2}{2t} \right)$$ the density of $B_t$. Then $$\Phi_t(x) = \mathbb{P}(B_t \leq x) = \int_{(-\infty,x]} p_t(y) \, dy.$$ Differentiation with respect to $t$ yields
$$\frac{d}{dt} \Phi_t(x) = \int_{(-\infty,x]} \frac{d}{dt} p_t(y) \, dy.$$ Since $$\frac{d}{dt} p_t(y) = \frac{1}{2} \frac{d^2}{dy^2} p_t(y)$$ we get $$\frac{d}{dt} \Phi_t(x) = \frac{1}{2} \int_{(-\infty,x]} \frac{d^2}{dy^2}p_t(y) \, dy = \frac{1}{2} \frac{d}{dx} p_t(x).\tag{1}$$
On the other hand, the fundamental theorem of calculus gives
$$\frac{d}{dx} \Phi_t(x) = p_t(x)$$
and so
$$\frac{d^2}{dx^2} \Phi_t(x) = \frac{d}{dx} p_t(x).\tag{2}$$
Combining $(1)$ and $(2)$ proves the assertion.