The probability in question is $1-p$, where $p$ is the probability that we will hit the negative $x$-semiaxis before hitting the positive $x$-semiaxis. Next, $p$ is the probability that (we will hit the positive $y$-semiaxis before hitting the positive $x$-semiaxis, and then we will hit the negative $x$-semiaxis before hitting the positive $x$-semiaxis). So, by the strong Markov property and the symmetry,
\begin{equation*}
p=q\,\tfrac12,
\end{equation*}
where $q$ is the probability that we will hit the positive $y$-semiaxis before hitting the positive $x$-semiaxis. In turn,
\begin{equation}
q=P(\tau_a<\tau_b)=\int_0^\infty P(\tau_a<t)\,dP(\tau_b<t), \tag{1}
\end{equation}
where $\tau_a$ and $\tau_b$ are independent random variables (r.v.'s) such that for any real $c>0$ the distribution of the r.v. $\tau_c$ is that of the time for a standard Brownian motion (starting at $0$) to first reach point $c$:
\begin{equation*}
P(\tau_c<t)=2(1-\Phi(c/\sqrt t))
\end{equation*}
for $t>0$, by the reflection principle, where $\Phi$ is the standard normal distribution function. Thus, by (1),
the probability in question is
\begin{equation}
1-q/2=
1-\int_0^\infty (1-\Phi(a/\sqrt t))\varphi(b/\sqrt t)\frac{b\,dt}{t^{3/2}}=1-\frac1\pi\,\arctan\frac ba,
\end{equation}
where $\varphi:=\Phi'$.

(One way to evaluate the latter integral is as follows: differentiate it in $a$, then use the substitution $t=x^{-2}$ to see that the derivative of the integral is $-\frac b{\pi(a^2+b^2)}$, and finally integrate this derivative back in $a$. Another way is to use the substitution $t=x^{-2}$ right away, to rewrite the integral as
\begin{equation*}
\int_0^\infty P(Z_1>ax)P(|Z_2|\in b\,dx)=P(Z_1>a|Z_2|/b)=P(C<b/a)=\frac1\pi\,\arctan\frac ba,
\end{equation*}
where $Z_1$ and $Z_2$ are independent standard normal r.v.'s, so that $C:=|Z_2|/Z_1$ is a Cauchy r.v.
Yet another way is suggested by Kostya.)