= Solution
Each marginal is a <Bessel process> of <dimension> $\delta=1+2a\in(1,2)$, but finite lifetime can also be proved directly. Apply the <Itô formula> before $X$ reaches zero:
$$
\log X_t=\log x+\int_0^tX_s^{-1}\,dB_s+
\left(a-\frac12\right)\int_0^tX_s^{-2}\,ds.
$$
With clock $u(t)=\int_0^tX_s^{-2}ds$, the <Itô integral> is a <Brownian motion> $W_u$. The inverse clock satisfies
$$
X_{t(u)}=x e^{W_u-(1/2-a)u},\qquad
t(u)=x^2\int_0^u e^{2W_v-(1-2a)v}\,dv.
$$
The clock runs through all $u\geq0$ up to the maximal lifetime: a finite terminal clock would leave $X$ at a finite positive limit, allowing continuation, or, at an infinite physical lifetime, would make its clock grow at a positive rate, a contradiction. There is no finite-time explosion to infinity; for example $dX_t^2=2X_t\,dB_t+(1+2a)dt$ and localization bounds the probability of reaching arbitrarily large levels on a fixed time interval. Since $W_u/u\to0$ almost surely and $1-2a>0$, the last exponential integral has finite limit, while $X_{t(u)}\to0$. Therefore
$$
\boxed{\zeta=x^2\int_0^\infty e^{2W_v-(1-2a)v}\,dv<\infty\quad\text{almost surely}.}
$$
The same argument applies to $Y$, whose noise $-B$ is also <Brownian motion>, so its lifetime is finite too. No <independence> between the two lifetimes is being claimed.
Set $S=X+Y$ and $U=Y/S$ before $T=\zeta\wedge\tau$. <Brownian motion> terms cancel in the sum:
$$
dS_t=a\left(\frac1{X_t}+\frac1{Y_t}\right)dt,
\qquad S_t\geq x+y>0.
$$
Thus simultaneous extinction is impossible. At the finite first lifetime, precisely one coordinate is zero and the other remains positive. Because $S$ has finite variation, differentiating the ratio gives
$$
dU_t=-\frac1{S_t}\,dB_t
+\frac{a(1-2U_t)}{S_t^2U_t(1-U_t)}dt.
$$
After clock change $v=\int_0^tS_s^{-2}ds$, its generator is
$$
\mathcal L=\frac12\frac{d^2}{du^2}
+\frac{a(1-2u)}{u(1-u)}\frac d{du}.
$$
A <scale function of a one-dimensional diffusion> $H$ solves $\mathcal LH=0$, so
$$
\frac{H''}{H'}=-2a\frac{1-2u}{u(1-u)},\qquad
H'(u)=C u^{-2a}(1-u)^{-2a}.
$$
Both endpoint singularities are integrable since $a<1/2$. Normalize $H(0)=0,H(1)=1$ to obtain
$$
H(u)=\frac{\displaystyle\int_0^u v^{-2a}(1-v)^{-2a}dv}
{\displaystyle\int_0^1 v^{-2a}(1-v)^{-2a}dv}.
$$
The <Itô formula> makes $H(U_{t\wedge T})$ a bounded <martingale>, first by localization away from the endpoints and then by <bounded convergence theorem>. Its terminal value is one on $\{\zeta<\tau\}$, when $U_T=1$, and zero on the opposite event, when $U_T=0$. <optional stopping theorem> therefore proves the <oppositely driven Bessel exit probability>
$$
\boxed{\mathbb P(\zeta<\tau)=
\frac1{\mathrm B(1-2a,1-2a)}\int_0^{y/(x+y)}\frac{du}{u^{2a}(1-u)^{2a}},
\qquad c_a=\frac{\Gamma(2-4a)}{\Gamma(1-2a)^2}.}
$$
Here $\mathrm B$ is the <beta function> and $\Gamma$ is the <Gamma function>.
\b[The integral printed in the question is incorrect for general $a$.] Its first exponent is $2-4a$, whereas the two coupled equations give $2a$. At $a=1/4$, which lies in the allowed interval, its integral from zero diverges for every positive upper limit; no finite normalizing constant gives a probability. Moreover exchanging $X$ and $Y$ while changing $B$ to $-B$ shows that at $x=y$ the probability must be $1/2$. The corrected symmetric integral has this value for every $a\in(0,1/2)$, whereas the printed asymmetric normalized integral generally does not. For $a>1/4$, even when the printed integral converges, its <derivative> density $H_{\mathrm{print}}'(u)\propto u^{-(2-4a)}(1-u)^{-2a}$ satisfies
$$
\frac{\mathcal LH_{\mathrm{print}}}{H_{\mathrm{print}}'}=\frac{3a-1}{u},
$$
so it is harmonic for the required exit problem only at $a=1/3$. The two exponents agree exactly at that value.
To relate that value to <percolation>, write the two <SLE> <boundary> gaps as $g_t(x)-\xi_t$ and $\xi_t-g_t(-y)$. In Brownian-time units $s=\kappa t$, they have the displayed coupled equations with $a=2/\kappa$, after reversing the <Brownian motion> sign. Hence $a=1/3$ corresponds to $\kappa=6$. The event is that the positive marked point $x$ is swallowed before the negative marked point $-y$.
For a quadrilateral with <boundary> points in the order $(-y,0,x,\infty)$, its conformal <cross-ratio> coordinate is $u=y/(x+y)$. In the critical <percolation> exploration, wire the arcs $(-y,0)$ and $(x,\infty)$ blue and the other two arcs yellow. Exploration from zero, stopped when one of the two marked sides is swallowed, detects a blue crossing between those two blue arcs exactly in the event just computed. The continuum exploration has the SLE6 law; its <Locality property of SLE> lets one stop the ordinary chordal exploration before the distant <boundary> colors affect it. Thus the crossing probability is
$$
\boxed{\frac{\Gamma(2/3)}{\Gamma(1/3)^2}\int_0^{y/(x+y)}[u(1-u)]^{-2/3}du,}
$$
the <Cardy boundary crossing formula>. This is the rigorous scaling-limit formula for critical <site percolation> on the triangular lattice; applying it to other critical planar models requires the corresponding universality assumption. At this <percolation> value of $a$, the printed integral is already the correct one.
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