Let $a>0$ be a real number, let $T=\mathbb R_{\ge0},$ let $(W(t))_{t\in T}=(W_1(t),W_2(t))_{t\in T}$ be a two-dimensional Brownian motion and let $\tau_a=\inf\{t\in T:W_2(t)=a\}.$ What is the distribution of $W_1(\tau_a)$?
From the Reflection Principle, it follows that $\tau_a$ follows a Lévy distribution with scale $c=a^2$ and location $\mu=0.$ In addition, we can show $\sigma(\tau_a)$ is independent of ${\cal G}:=\sigma(W_1(t),t\in T).$ We can also show that for any Borel bounded function $f:\mathbb R\to\mathbb R,$ we have $$\mathbb E(f(W_1(\tau_a))\mid\sigma(\tau_a))=\frac1{\sqrt{2\pi\tau_a}}\int_{\mathbb R}f(x)e^{-x^2/(2\tau_a)}dx,$$ by verifying the defining property of conditional expectation, first for the case $f=I_B$ is an indicator of a Borel set and then for general Borel bounded $f$ by using the Functional Monotone Class Theorem. I was thinking that the above result maybe holds for $f(x)=x,$ so that if $W_1(\tau_a)\in L^1$ then $$\mathbb E(W_1(\tau_a))=\mathbb E\mathbb E(W_1\circ\tau_a\mid\sigma(\tau_a))=\mathbb E\left(\frac1{\sqrt{2\pi\tau_a}}\int_{\mathbb R}xe^{-x^2/2\tau_a}dx\right)=0,$$ where the last equality follows from $\forall b\in\mathbb R_{>0}\quad\int_{-\infty}^\infty xe^{-x^2/(2x)}dx=0.$ And if the result holds for $f(x)=x^2$ and if $W_1(\tau_a)\in L^2,$ then from $\forall b\in\mathbb R_{>0}\quad\int_{-\infty}^\infty x^2e^{-x^2/2a}dx=\sqrt{2\pi b^3}$ we similarly have $$\mathbb E(W_1(\tau_a)^2)=\mathbb E\left(\frac1{\sqrt{2\pi\tau_a}}\int_{\mathbb R}x^2e^{-x^2/(2\tau_a)}dx\right)=\mathbb E\left(\frac1{\sqrt{2\pi\tau_a}}\sqrt{2\pi\tau_a^3}\right)$$ $$=\mathbb E\tau_a=\infty,$$ so that $W_1(\tau_a)\not\in L^2.$ I don't know what to do from here.
Okey, let $W=(W_1,W_2)$ be a $2$-dimensional Wiener Process. Let $\tau_a = \inf\{t > 0: W_2(t) = a\}$ and let $Z:= W_1(\tau_a)$ (that is $Z(\omega) = W_1(\tau_a(\omega))(\omega)$). We want to calculate the distribution of random variable $Z$. As you noted, since $\tau_a$ is $\mathcal F^{W_2}_\infty$ measurable it is independent of process $W_1$. Note that $Z=F((W_1),\tau_a)$ where $F:C([0,\infty))\times \mathbb R_+ \to \mathbb R$ is given by $F(g,x) = g(x)$. You can check it is a measurable function. By that we get:
$\mathbb P(Z \le t) = \mathbb P(F((W_1),\tau_a) \le t) = \mathbb E[ \mathbb P(F((W_1),\tau_a)) \le t | \tau_a)]$
By properties of conditional expectation, since $W_1,\tau_a$ are independent, we get $\mathbb P(F((W_1),\tau_a)) \le t | \tau_a) = \mathbb P(F((W_1),p) \le t)|_{p = \tau_a}$.
So let's calculate $\mathbb P( W_1(p) \le t ) = \mathbb P( \sqrt{p}W_1(1) \le t) = \mathbb P(W_1(1) \le \frac{t}{\sqrt{p}}) = \Phi(\frac{t}{\sqrt{p}})$, where $\Phi$ is CDF of $\mathcal N(0,1)$ random variable.
By that we get: $\mathbb P(Z \le t) = \mathbb E [ \Phi(\frac{t}{\sqrt{\tau_a}})] $
Recall that $\tau_a$ has density $\frac{1}{\sqrt{2\pi}}\frac{a}{\sqrt{y^3}}\exp(-\frac{a^2}{2y})1_{(0,\infty)}(y) =: g(y)$, where $1_A$ stands for indicator function.
Plugging it in: $\mathbb P(Z \le t) = \int_\mathbb R \Phi(\frac{t}{\sqrt{y}})g(y) dy$. It can be shown, that the derivative of the function under integral is well defined for every $t$ and is integrable with respect to $y$ (you can try to prove that we can go with derivative under integral sign. It is just to apply dominated convergence, cause we get a bound from above by integrable function by Lagrange mean value theorem), we get that density of $Z$ is given by:
$h(t) = \int_{\mathbb R} \frac{1}{\sqrt{y}}g(y) \phi(\frac{t}{\sqrt{y}})dy$, where $\phi$ is density of standard normal. Plugging everything in, we finally arrive at: (hopefully I didn't miss anything here)
$$ h(t) = \frac{a}{2\pi}\int_0^\infty \frac{1}{y^2}\exp(-\frac{a^2}{2y}) \exp(-\frac{t^2}{2y})dy $$
Take substitution $z=\frac{1}{y}$ getting $dz = \frac{-1}{y^2}dy$ and $y=0 \rightarrow z=\infty$, $y = \infty \rightarrow z = 0$ we get:
$$ h(t) = \frac{a}{2\pi} \int_0^\infty \exp(-\frac{a^2 z}{2})\exp(-\frac{t^2 z}{2})dz = \frac{a}{2\pi} \int_0^\infty \exp(-z( \frac{a^2+t^2}{2}))dz$$ so again, if I've done calculations correct, we should get:
$$ h(t) = \frac{a}{2\pi} \frac{2}{a^2+t^2} = \frac{a}{\pi(a^2+t^2)} $$ so it seems that $W_1(\tau_a)$ is distributed as $Cauchy(a)$. As I see, you stated that $W_1(\tau_a) \in L_1$, but that wasn't quite proper way to do it. To use conditional expectation, you must be sure beforehand that this variable is in $L_1$. So you cant do $\mathbb E[X] = \mathbb E \mathbb E[X|Y]$ if $X$ is not integrable, so you can't use it to show that $X$ is integrable. I conditioned, too, that is right, but I conditioned $\mathbb P( W_1(\tau_a)) \le t) = \mathbb E[1_{\{W_1(\tau_a) \le t \}}]$ which is bounded, so obviously that indicator function is integrable.
Just some thoughts on the margin, it is quite interesting, since Normal distribution is $2-$stable (that means $aX + bY \sim \sqrt{a^2+b^2}\mathcal N(0,1)$ when $X,Y$ are i.i.d $\mathcal N(0,1)$), and it can be shown that $\tau_{(.)}$ distribution is $\frac{1}{2}-$ stable (that means $aX + bY = (\sqrt{a} + \sqrt{b})^2 \tau_1$, where $X,Y$ are independent distributed as $\tau_1$). It won't be that interesting, but Cauchy distribution, is $1-$stable. So somehow by combining $2-$ stable and $\frac{1}{2}-$stable in proper way you get $1-$stable (Of course, if my calculations are correct, but I hope and believe so)