= Solution
Let $y_n=\langle e_n,y\rangle$. Using the phase of $u_n$ in the <singular value system>, the <Moore–Penrose inverse of an operator> becomes
$$
\boxed{f^\dagger=\sum_{n\in\mathbb Z}\frac{y_n}{c_n}e_n,\qquad\sum_{n\in\mathbb Z}\frac{|y_n|^2}{|c_n|^2}<\infty.}
$$
This gives the <admissible data for a periodic convolution inverse>. Because the range is dense and the kernel is zero here, the domain of the inverse is exactly the range, not all of $L^2$.
For $y(x)=e^{\alpha x}$ and $\alpha\notin i\mathbb Z$,
$$
y_n=\frac1{\sqrt{2\pi}}\int_0^{2\pi}e^{(\alpha-in)x}dx=\frac{e^{2\pi\alpha}-1}{\sqrt{2\pi}(\alpha-in)}.
$$
The formal expression would consequently be
$$
f_{\rm formal}(x)=\frac{e^{2\pi\alpha}-1}{2\pi}\sum_{n\in\mathbb Z}\frac{e^{inx}}{(\alpha-in)c_n}.
$$
However, the <high-frequency obstruction for nonperiodic exponential data> prevents this from being a Hilbert-space solution. The numerator is nonzero, $|y_n|$ is asymptotic to a nonzero constant divided by $|n|$, and part (c) proved $|c_n|=o(1/|n|)$. Hence $|y_n/c_n|\to\infty$, so the <Picard criterion> fails. \b[For $\alpha\notin i\mathbb Z$, $A^\dagger y$ is not defined in the specified <Hilbert space>.] The formal series is not a convergent generalized solution.
If $\alpha=im$ with $m\in\mathbb Z$, the data are a single periodic Fourier mode: $y_n=\sqrt{2\pi}\delta_{nm}$. Then there is an exact unique solution,
$$
\boxed{f^\dagger(x)=\frac{e^{imx}}{c_m}\quad\text{when }\alpha=im.}
$$
In particular, if the intended parameter is real, only $\alpha=0$ is admissible, giving $f^\dagger=1/c_0$.
For the inadmissible cases the inverse problem still has approximate solutions $f_N=\sum_{|n|\leq N}(y_n/c_n)e_n$: their images are Fourier projections converging to $y$ in $L^2$, while their <norms> diverge. Thus the least-squares residual has infimum zero but no minimizer. This distinguishes an undefined exact inverse from a regularized truncated reconstruction; it is the necessary qualification to the question's unrestricted constant $\alpha$.
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