= Solution
Set $u_0=\alpha/\beta$ and use the accelerated grid time $j/n$. At population $k=nu_0+\sqrt n\,x$, the fluctuation increment is $h_n=n^{-1/2}$, $-h_n$, or zero. Its first and second conditional moments, multiplied by $n$, are
$$
b_n(x)=n\mathbb E(\Delta X^n\mid x)
=\sqrt n\,(p_k-q_k)=-\alpha x-\frac\beta{\sqrt n}x^2,
$$
$$
a_n(x)=n\mathbb E((\Delta X^n)^2\mid x)
=p_k+q_k=\frac{2\alpha^2}{\beta}
+\frac{3\alpha}{\sqrt n}x+\frac\beta n x^2.
$$
Thus $b_n\to b(x)=-\alpha x$ and $a_n\to a=2\alpha^2/\beta$, uniformly on every compact set. The centered conditional variance has the same limit, since the squared conditional mean contributes only $b_n(x)^2/n$.
Here is the <diffusion approximation theorem> being applied. For Markov chains sampled with time step $1/n$, suppose the first and second conditional increment moments multiplied by $n$ converge uniformly on compact sets to continuous coefficients $b,a$; the large-jump conditional second moments satisfy the local Lindeberg condition; the initial laws converge; compact containment holds on finite time intervals; and the limiting <martingale problem> is well-posed and nonexplosive. Then the interpolated chains converge weakly, uniformly on finite time intervals, to that problem's continuous solution. The step versions converge in the <Skorokhod J1 topology>. Local moment bounds supply the stopped-process tightness, and the limiting test-function identities identify every subsequential limit by the <martingale problem>.
We verify the global condition rather than assume it. Before the stopping rule is triggered, $p_k+q_k\le1$, and $k\ge0$ implies $x\ge-u_0\sqrt n$. Therefore
$$
xb_n(x)=-x^2\left(\alpha+\frac\beta{\sqrt n}x\right)\le0.
$$
For the <Lyapunov function> $V(x)=x^2$, the exact one-step generator is consequently
$$
G_nV(x)=2xb_n(x)+a_n(x)\le1.
$$
After stopping, the process is constant. Applying the elementary stopped <supermartingale> inequality for $V(X_{j/n}^n)-j/n$, and stopping also on first reaching $|X^n|\ge L$, gives
$$
\boxed{\mathbb P\left(\sup_{t\le1}|X^n_{t\wedge\tau_n}|\ge L\right)
\le\frac{(X_0^n)^2+1}{L^2}.}
$$
This is <compact containment>. At zero population we use the natural absorbing convention $p_0=q_0=0$; the same calculation is valid there.
The invalid-probability threshold is at the positive root
$$
u_* =\frac{\sqrt{\alpha^2+4\beta}-\alpha}{2\beta}
\quad\text{of}\quad\alpha u+\beta u^2=1.
$$
Because $2\alpha^2<\beta$, we have $u_*>u_0$. The corresponding fluctuation level $\sqrt n(u_*-u_0)$ tends to infinity, so the stopping rule changes none of the compact-set moment limits. The initial fluctuation satisfies $|X_0^n|\le n^{-1/2}$ and hence tends to zero. All jumps have magnitude at most $n^{-1/2}$, so the Lindeberg condition is immediate. The limiting equation has globally Lipschitz coefficients and is well-posed and nonexplosive. The theorem therefore gives
$$
\boxed{dX_t=-\alpha X_tdt+\alpha\sqrt{\frac2\beta}\,dB_t,
\qquad X_0=0.}
$$
This is the <Ornstein-Uhlenbeck fluctuation limit at a stable population equilibrium>. The limiting <Ornstein-Uhlenbeck process> has explicit solution
$$
X_t=\alpha\sqrt{\frac2\beta}\int_0^te^{-\alpha(t-s)}dB_s,
\qquad \operatorname{Var}(X_t)=\frac\alpha\beta(1-e^{-2\alpha t}).
$$
The PDF samples the linearly interpolated population at $\lfloor nt\rfloor$, so its displayed process is actually the step version. Using $nt$ instead gives the continuous interpolation; their uniform distance is at most $n^{-1/2}$. Both therefore have the asserted continuous diffusion limit, with the step version interpreted in <Skorokhod J1 topology>.
Finally, if $\tau_n\le1$, the stopped fluctuation reaches a value greater than $\sqrt n(u_*-u_0)$. The preceding <Lyapunov function> bound gives directly
$$
\boxed{\mathbb P(\tau_n\le1)
\le\frac{(X_0^n)^2+1}{n(u_*-u_0)^2}\longrightarrow0.}
$$
This also shows explicitly why the artificial stopping rule is asymptotically irrelevant.
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