= Solution
The exponential map is a conformal bijection from the strip
$$
\mathcal S=\{z:-\alpha<\operatorname{Im}z<\beta\}
$$
to the wedge containing the starting point $1$. The restriction $\alpha+\beta<2\pi$ makes it injective. A <planar Brownian motion> $U+iV$ started at zero in this strip maps, after the <conformal Brownian clock>, to the required <Brownian motion> in the wedge.
Let $\sigma$ be the first time $V$ exits $(-\alpha,\beta)$. It is finite <almost surely>, and $V_{t\wedge\sigma}$ is a bounded <martingale>. The <optional stopping theorem> gives
$$
0=\mathbb E V_\sigma
=\beta p-\alpha(1-p),
\qquad p=\mathbb P(V_\sigma=\beta).
$$
The strip exit maps to the corresponding bounding ray. The exponential clock is finite at $\sigma$, because its integrand is continuous on the compact time interval $[0,\sigma]$. Thus it does not remove or reverse the exit event. We obtain
$$
\boxed{\mathbb P(S_\beta<S_{-\alpha})=\frac{\alpha}{\alpha+\beta}.}
$$
This is the <Brownian wedge-exit probability> obtained directly from the strip, using <conformal invariance of planar Brownian motion>.
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