For $h:=\Delta t>0$, you had 
$$P(\tau>t)-P(\tau>t+h)=\int_{(0,\infty)}
P(M\le-x|X_t=x)\,P(X_t\in dx),$$
where 
$$M:=\inf_{t\le u\le t+h}J_u,\quad J_u:=\int_t^u a(s,X_s)\,dW_s.$$
By [Doob's martingale inequality][1], 
$$P(M\le-x|X_t=x)\le x^{-2}\,E(J_{t+h}^2|X_t=x)
\le x^{-2}\,ha_2^2,$$
where $a_2:=\overline a$ and $a_1:=\underline a$.

The crucial point is that the pdf $p_t$ of $X_t$ for $t>0$ is [bounded][2] so that 
$$p_t(x)\le\frac c{\sqrt t}\,e^{-bx^2/t}\le\frac c{\sqrt t}$$
for some positive real constants $c,b$ depending only on $a_1,a_2,L$ and for all real $x$. So, 
$$
\begin{align}
P(\tau>t)-P(\tau>t+h)&\le
\int_{(0,\infty)}\min(1,x^{-2}\,ha_2^2)\,\frac c{\sqrt t}\,dx \\ 
&=
\sqrt h\int_{(0,\infty)}\min(1,u^{-2}\,a_2^2)\,\frac c{\sqrt t}\,du \\ 
&=\frac C{\sqrt t}\,\sqrt h,
\end{align}
$$ 
where $C>0$ is a real constant depending only on $a_1,a_2,L$. 

---

To get now a uniform Hölder continuity, we can reason as follows: 
$$P(\tau>t)-P(\tau>t+h)\le1-P(\tau>t+h)=P(\tau\le t+h)\le a_2^2(t+h)/1^2,$$ 
again by Doob's martingale inequality. So, 
$$P(\tau>t)-P(\tau>t+h)\le\min\Big(1,\frac C{\sqrt t}\,\sqrt h,a_2^2(t+h)\Big)\le C_1h^{1/3},$$ 
where $C_1>0$ is a real constant depending only on $a_1,a_2,L$ (to verify the latter inequality, consider separately the three cases when (i) $h\ge1$, (ii) $t\le h^{1/3}<1$, or (iii) $t>h^{1/3}$). 

Using here an exponential inequality (see e.g. [Theorem 3.1][3]) instead of Doob's one, one can improve the factor $h^{1/3}$ to  $h^{1/2}\ln\frac1h.$ 


  [1]: https://en.wikipedia.org/wiki/Doob%27s_martingale_inequality#Statement_of_the_inequality
  [2]: https://mathoverflow.net/a/405686/36721
  [3]: https://projecteuclid.org/download/pdf_1/euclid.aop/1176988477