I've found this paper by AitSahlia and Lai (2001) talking about the change in variables they use in order to begin solving the American Put Integral equation where they go from here:
\begin{split} K-\pi_t = & \; P_f^E(\pi_t, t) \\ & \; + \int_t^T \Big[ rKe^{-r(\tau-t)} N(-d_2(\pi_t, \pi_\tau, \tau-t))\Big]\;\mathrm{d}\tau \\ & \; - \int_t^T \Big[\delta e^{-\delta(\tau-t)} N(-d_1(\pi_t, \pi_\tau, \tau-t)) \Big]\;\mathrm{d}\tau \end{split}
where:
\begin{split} P_f^E(S,K,r, \delta, \sigma, T) = & \; Ke^{-r(T-t)}N(-d_2(S,K,T-t))-SN(-d_1(S,K,T-t)) \\ d_1(x,y,z)= & \; \dfrac{ln(\frac{x}{y})+(r-\delta + \frac{\sigma^2}{2})z}{\sigma\sqrt{z}} \\ d_2(x,y,z)= & \; d_1(x,y,z)-\sigma\sqrt{z} \\ \end{split}
and with the assumptions:
\begin{split} \rho = \dfrac{r}{\sigma^2} & \; \qquad \alpha = \dfrac{\delta}{r} \\ s=\sigma^2(t-T) \qquad & \; z=\log(S/K)-(\rho-\alpha\rho-\frac{1}{2})s \end{split}
yield the outcome:
\begin{split} \pi_t = Ke^{\tilde{z}(s)+(\rho-\alpha\rho-\frac{1}{2})s} \end{split}
where, and I quote, "boundary $\pi_t$ becomes $\tilde{z}(s)$ in the new co-ordinate system, with $\pi_t$" equal to the above . Maybe it's just me but the grammar seems horrible on this line and I don't know what it's trying to say regarding the new outcome. I've also tried to substitute back in the variables to $z$ for which I get somewhere close to $d_1$ but not equal.
If anyone could point me in the direction of an explanation of what happens here or could explain it themselves in layman terms, that would be great.