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Let $W_t= -B_t$. Moreover, let $a= - \frac{r-\frac{1}{2}\sigma^2}{\sigma}$ and $b= -\frac{1}{\sigma}\ln \frac{S^*}{S_0}$. Then, as in this question, \begin{align*} \mathbb{P}\left(\tau \ge T \mid W_T\right)\pmb{1}_{\{W_T \le b-aT\}} &= \mathbb{P}\left(W_t+at \le b, t\in[0, T] \mid W_T\right)\pmb{1}_{\{W_T \le b-aT\}}\\ &=\Big[1-\exp\Big(-\frac{2}{T}b\big(b-W_T-aT\big)\Big)\Big] \pmb{1}_{\{W_T \le b-aT\}}, \end{align*} and, consequently, \begin{align*} \mathbb{P}(\tau \ge t) &= \Phi\left(\frac{b-at}{\sqrt{t}}\right) - e^{2ab}\Phi\left(\frac{-b-at}{\sqrt{t}}\right), \end{align*} where $\Phi$ is the cumulative distribution function of a standard normal random variable. The density function is then given by \begin{align*} \frac{b}{\sqrt{2\pi t^3}}e^{ab}e^{-\frac{1}{2}\left(\frac{b^2}{t}+a^2t \right)}\pmb{1}_{t> 0} = \frac{b}{\sqrt{2\pi t^3}} e^{-\frac{1}{2}\left(\frac{b-at}{\sqrt{t}}\right)^2}\pmb{1}_{t> 0} \end{align*} Therefore, \begin{align*} E\left( 1_{\tau \leq \infty }e^{-r\tau}\right) &= \int_0^{\infty} e^{-rt}\frac{b}{\sqrt{2\pi t^3}} e^{-\frac{1}{2}\left(\frac{b-at}{\sqrt{t}}\right)^2}dt \\ &=e^{ab - b \sqrt{2r+a^2}}\int_0^{\infty} \frac{b}{\sqrt{2\pi t^3}} e^{-\frac{1}{2}\left(\frac{b-\sqrt{2r+a^2}t}{\sqrt{t}}\right)^2}dt \\ &=e^{ab - b \sqrt{2r+a^2}}\left[\Phi\left(\frac{b-\sqrt{2r+a^2}t}{\sqrt{t}}\right) - e^{2b\sqrt{2r+a^2}}\Phi\left(\frac{-b-\sqrt{2r+a^2}t}{\sqrt{t}}\right) \right]_{\infty}^0\\ &=e^{ab - b \sqrt{2r+a^2}}. \end{align*}