You can indeed get the distribution of first passage times of the Wiener process in this way, but it's a bit more effort than the standard way. All references in this answer are to Analytic Combinatorics, which you can download for free.
As Robert showed, the probability generating function for the first passage time $\tau(n)$ to $n$ is
$g_n(s) = \left(\frac{1-\sqrt{1-s^2}}{s}\right)^n\;.$
I'm using $n$ instead of $m$ to match the variables in Theorem VIII.8 (p. 587). First we need to bring this into a form in which we can apply the theorem, that is, we need to satisfy conditions $\mathbf L_1$ and $\mathbf L_2$ in Section VIII.8.1 (p. 586):
$g_n(s)=s^n\left(\frac{1-\sqrt{1-s^2}}{s^2}\right)^n=:s^nB(s^2)^n$
with
$B(z)=\frac{1-\sqrt{1-z}}z\;.$
To find the distribution of the first passage time $t$ at $a$, we need to take the step sizes $\Delta x$ and $\Delta t$ to zero while keeping $\Delta x^2/\Delta t=:\beta$ constant. Then $n=a/\Delta x$ and $N=t/(2\Delta t)=\beta t/(2\Delta x^2)$, where the factor $2$ arises from $z=s^2$. We need to find the asymptotic behaviour of the coefficients of $B(z)$ as $\Delta x$ goes to zero and thus both $n$ and $N$ go to infinity.
First note that the spread $T$ defined in Section VIII.8.1 is infinite in this case, so we don't have to worry about restrictions on the ratio $\lambda=N/n=\beta t/(2a\Delta x)$, which goes to infinity as $\Delta x\to0$. We have to find the root $\zeta$ of
$\zeta\frac{B'(\zeta)}{B(\zeta)}=\lambda\;,$
which turns out to have the nice form
$\zeta=1-\frac1{(2\lambda+1)^2}$
with
$B(\zeta)=1-\frac1{2(\lambda+1)}\;.$
We also need
$\xi=\frac{\mathrm d^2}{\mathrm d\zeta^2}\left(\log B(\zeta)-\lambda\log\zeta\right)=\frac{B''(\zeta)}{B(\zeta)}+O\left(\lambda^2\right)=(2\lambda)^3+O\left(\lambda^2\right)\;.$
Now we can apply Theorem VIII.8 to obtain the distribution function as the limit
$ \begin{align} \lim_{\Delta x\to0}\frac{[s^{2N}]g_n(s)}{\Delta t} &= \lim_{\Delta x\to0}\frac{[z^{N-n/2}]B(z)^n}{\Delta t} \\ &= \lim_{\Delta x\to0}\frac{[z^N]B(z)^n}{\Delta t} \\ &= \lim_{\Delta x\to0}\frac{B(\zeta)^n}{\zeta^{N+1}\sqrt{2\pi n\xi}\Delta t} \\ &= \lim_{\Delta x\to0}\frac{\left(1-\dfrac1{2(\lambda+1)}\right)^n}{\left(1-\dfrac1{(2\lambda+1)^2}\right)^{N+1}\sqrt{2\pi (2\lambda)^3\dfrac a{\Delta x}}\,\Delta t} \\ &= \lim_{\Delta x\to0}\frac{\left(1-\dfrac{a\Delta x}{\beta t}\right)^{a/\Delta x}}{\left(1-\left(\dfrac{a\Delta x}{\beta t}\right)^2\right)^{\beta t/2(\Delta x^2)+1}\sqrt{2\pi\left(\dfrac{\beta t}{a\Delta x}\right)^3\dfrac a {\Delta x}}\,\Delta x^2/\beta} \\ &= \frac{\exp\left(-\dfrac{a^2}{\beta t}\right)}{\exp\left(-\dfrac{a^2}{2\beta t}\right)\sqrt{\dfrac{2\pi\beta t^3}{a^2}}} \\ &= \frac{a\exp\left(-\dfrac{a^2}{2\beta t}\right)}{\sqrt{2\pi\beta t^3}}\;. \end{align} $
Then setting $\beta=1$ gives the desired result; it appears you forgot a square root around $2\pi$.