It's not difficult to see that
$X_t := \exp \left(\sqrt{2} B_t \right)$
solves the given SDE. (You can either use Itô's formula to check it or use some standard methods for linear SDE's to obtain this solution.) Moreover, by Ito's formula:
$f(t,X_t)-\underbrace{f(0,x_0)}_{x_0} = \sqrt{2} \int_0^t e^{-s} \cdot X_s \, dB_s + \int_0^t e^{-s} \cdot X_s + (-e^{-s} \cdot X_s) \, ds \\ = \sqrt{2} \int_0^t e^{-s} \cdot e^{\sqrt{2} B_s} \,dB_s \\ \Rightarrow f(t,X_t) = \underbrace{\sqrt{2} \int_0^t e^{\sqrt{2} B_s-s} \,dB_s}_{M_t} + \underbrace{x_0}_{A_t}$
where $x_0=1$. Let
$g(s,w) := \sqrt{2} \cdot e^{\sqrt{2} B(s,w)-s}$
Then $g \in L^2(\lambda_T \otimes \mathbb{P})$, i.e.
$\int_0^T \int_\Omega g(s,w)^2 \, d\mathbb{P} \, ds <\infty$
There is a general result which says that this condition implies that $M_t$ is a martingale (and not only a local one). Moreover,
$\langle M,M \rangle_t = \int_0^t |g(s,w)|^2 \, ds$
(see René L. Schilling/Lothar Partzsch: "Brownian Motion - An Introduction to stochastic processes", Theorem 14.13).
Concerning the integral $\mathbb{E}(e^{-\tau} \cdot X_\tau)$: Remark that
$\tau = \inf\{t \geq 0; X_t=2-t\} = \inf\{t \geq 0; \sqrt{2} B_t = \ln(2-t)\}$
Now let
$\sigma := 2\tau = \inf\{t \geq 0; \underbrace{\sqrt{2} B_{\frac{t}{2}}}_{=:W_t} = \ln (2-t/2)\}$
where $(W_t)_{t \geq 0}$ is again a Brownian Motion (scaling property). Thus
$\mathbb{E}(e^{-\tau} \cdot X_\tau) = \mathbb{E}(e^{-\tau+\sqrt{2} B_\tau}) = \mathbb{E}(e^{-\frac{\sigma}{2}+W_\sigma}) \stackrel{\ast}{=} 1$
In $(\ast)$ we applied the exponential Wald identity (see remark).
Remark Exponential Wald identity: Let $(W_t)_{t \geq 0}$ a Brownian motion and $\sigma$ a $\mathcal{F}_t^W$-stopping time such that $\mathbb{E}e^{\sigma/2}<\infty$, then $\mathbb{E}(e^{W_\sigma-\frac{\sigma}{2}})=1$. (see René L. Schilling/Lothar Partzsch: "Brownian Motion - An Introduction to stochastic processes", Theorem 5.14)