Since $B_t \to \infty$ a.s. as $t \to \infty$, we conclude from the intermediate value theorem that $T<\infty$ a.s. Moreover, by the continuity of the sample paths,
$$\mathbb{E}B_T = \mathbb{E}(e^{-\lambda T}) \neq 0$$
and therefore Wald's identities imply that $T$ is not integrable (if $T$ would be integrable, then $\mathbb{E}B_T = 0$).
Concerning the Ornstein-Uhlenbeck process: Let
$$X_t = \sigma \cdot e^{b \, t} \cdot \int_0^t e^{-b \, s} \, dB_s$$
where $\sigma>0$. Then
$$X_t \geq e^{-\lambda \, t} \Leftrightarrow M_t := \int_0^t e^{-b \, s} \, dB_s \geq \frac{1}{\sigma} e^{-(\lambda+b) \, t}$$
$M$ is a martingale, $M_0 =0$. A similar calculation as in the proof of Wald's identity shows that $\mathbb{E}M_T = 0$ for any integrable stopping time $T \in L^1$. With the same reasoning as above, we conclude $T \notin L^1$.
(This is far from being a complete answer to your question... but it's too long for a comment.)
Your question is related to subordination of Lévy processes. Before explaining this in more detail, let me rephrase the result which you stated.
Theorem 1: Let $(B_t)_{t \geq 0}$ and $(W_t)_{t \geq 0}$ be independent one-dimensional Brownian motions. If $$T_{t} := \inf\{s>0; W_s > t\} $$ is the first hitting time of $(t,\infty)$, then the process $$L_t := B_{T_t}, \qquad t \geq 0, $$ is a Cauchy process.
Proof: Because of the independence of $(T_t)_{t \geq 0}$ and $(B_t)_{t \geq 0}$, we have
$$\mathbb{P}(L_{t} \in A) = \int_{(0,\infty)} \mathbb{P}(B_s \in A) \, d\mathbb{P}_{T_t}(s)$$ Now we can plug in the distributions of $B_s$ and $T_t$ to conclude that $$\mathbb{P}(L_t \in A) = \int_A \left( \int_0^{\infty} p(s,x)f_{T_t}(s) \, ds \right) \, dx = \int_A \frac{t}{\pi (t^2+|x|^2)} \, dx$$ which shows that $L_t$ is Cauchy distributed for each $t$.
It is possible to show that the process $(T_t)_{t \geq 0}$ is a Lévy process (it has stationary and independent increments) with increasing sample paths, a so-called subordinator. There is the following general statement which states that Lévy processes are closed under subordination.
Theorem 2: Let $(X_t)_{t \geq 0}$ be a Lévy process and $(S_t)_{t \geq 0}$ an independent subordinator. Then the subordinated process $$L_t := X_{S_t}$$ is again a Lévy process.
The Brownian motion is a (very particular) case of a Lévy process. We can read Theorem 1 as follows: If we subordinate the Brownian motion $(B_t)_{t \geq 0}$ with the independent subordinator $(T_t)_{t \geq 0}$, then we get a Cauchy process.
It is natural to ask what happens if we replace the Brownian motions $(B_t)_{t \geq 0}$ and $(W_t)_{t \geq 0}$ by general Lévy processes. It follows from Theorem 2 that the resulting process will be a Lévy process, but it is, in general, very hard to identify its distribution. One of the few cases where this is possible is the following one:
If we replace $W_t$ in Theorem 1 by a scaled Brownian motion with drift $\delta^{-1}(W_t+\gamma t)$, then the subordinated process $L_t$ has a normal inverse Gaussian distribution.
More precisely, if $X$ ia a normal inverse Gaussian random variable with mean $0$, then its density function can be written in the form
$$\int_0^{\infty} f_{T_t}(s) p(x,s) \, ds$$
where $p$ is the transition density of Brownian motion, and $f_{T_t}$ the density of the first hitting time of $\delta^{-1}(W_t+\gamma t)$ (for suitable constants $\delta,\gamma$), see e.g. (1) for details.
(1) O.E. Barndorff-Nielsen: Normal Inverse Gaussian Distributions and Stochastic Volatility Modelling. Scandinavian Journal of Statistics 24 (1997), 1-13.
Best Answer
"Is the Cauchy density function the probability function of the second Brownian motion at time $\tau_a$, when the first Brownian motion is stopped?" Yes.
"The probability of the a Brownian motion should only have a growing variance, while the mean remains zero. I do not understand the Cauchy density distribution in this case." As $a$ increases, so will $\tau_a$ overall; and although a Cauchy distribution doesn't have finite variance, the distribution does get more dispersed as $a$ increases. So there doesn't seem to be a conflict with your understanding that the variance of Brownian motion is increasing in time.
And the solution to the problem: \begin{align} f_{W_2(\tau_a)}(y)dy &=P[W_2(\tau_a)\in(y,y+dy)]\\ &=\int_{t=0}^\infty P[W_2(\tau_a)\in(y,y+dy)|\tau_a=t]f_{\tau_a}(t)dt\\ &=\int_{t=0}^\infty \frac{1}{\sqrt{2\pi t}}e^{-\frac{y^2}{2t}}dy\frac{1}{\sqrt{2\pi t^3}} ae^{-\frac{a^2}{2t}}dt\\ &= \left(\int_{0}^\infty \frac{a}{2\pi t^2}e^{-\frac{y^2+a^2}{2t}}dt\right)dy,\quad\text{let }s=\frac{1}{t},\\ &= \left(\int_\infty^0\frac{as^2}{2\pi}e^{-\frac{y^2+a^2}{2}s}\left(-\frac{1}{s^2}\right)ds \right)dy\\ &=\left( \int_0^\infty\frac{a}{2\pi}e^{-\frac{y^2+a^2}{2}s}ds \right)dy\\ & = \frac{a}{\pi(a^2+y^2)}dy. \end{align}