My complex analysis is very sketchy, and I am a little stumped by the following - although it seems incredibly innocuous.
For $t\in\mathbb R$ and a fixed parameter $\alpha\in\mathbb R/\{0\}$ does it ever follow, or make sense to say, that $e^{-\alpha t}\in\mathbb C$?
My initial thought is that no, this can't be the case, since we have the image of real number under the exponential. However, I am looking at a problem which has that,
$$\langle e^{-\alpha t}x,y\rangle_{\mathcal H}=e^{-\alpha t}\langle x,y\rangle_{\mathcal H}=\langle x,e^{\alpha t}y\rangle_{\mathcal H},$$
where $x,y\in\mathcal H$, a Hilbert space, over the field $\mathbb K\in\{\mathbb R,\mathbb C\}$.
The only way that I can see that this would make sense would be if $e^{-\alpha t}\in\mathbb C$, but can this be justified with the information given?
I don't think "purely complex number" is a standard terminology (as opposed to "purely imaginary"). What you could say is "nonreal complex number".
Anyway, your computation is incorrect. If $\alpha,t$ are real (and $\langle\cdot,\cdot\rangle$ is an inner product), then $\langle e^{-\alpha t}x,y\rangle=\langle x,e^{-\alpha t}y\rangle$. What you wrote would be correct only if $\lvert e^{-\alpha t}\rvert=1$, or equivalently, if $\alpha t$ is purely imaginary --- only in this case it is true that $(e^{-\alpha t})^{-1}=\overline{e^{-\alpha t}}$.
My guess is that wherever you are taking this from, there's a missing imaginary unit, i.e., it should be $\langle e^{-\alpha ti}x,y\rangle=\langle x,e^{\alpha t i}y\rangle$.