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 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 $a_1,a_2,L$. Using here an exponential inequality instead of Doob's one, one can improve $h^{1/3}$ 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