In the usual definition of the (bilateral) $\mathcal{Z}$-transform, the series
$$X(z)=\sum_{n=-\infty}^{\infty}x[n]z^{-n}\tag{1}$$
converges in an annulus $r_1<|z|<r_2$.
Similarly, the integral of the (bilateral) Laplace transform
$$X(s)=\int_{-\infty}^{\infty}x(t)e^{-st}dt\tag{2}$$
converges in the vertical strip $\sigma_1<\textrm{Re}\{s\}<\sigma_2$.
There are certain sequences $x[n]$ and functions $x(t)$ for which the regions of convergence of $(1)$ and $(2)$ are reduced to a circle or a line, respectively. In these cases one would say that the $\mathcal{Z}$-transform or the Laplace transform don't exist. If the circle where the sum converges is the unit circle (or if the line where the integral converges is the imaginary axis), we can say that the (discrete-time or continuous-time) Fourier transforms exist.
Examples of such cases are the sequence
$$x[n]=\frac{\sin(\alpha n)}{\pi n}\tag{3}$$
for which the sum $(1)$ only converges for $|z|=1$.
Similarly, for the function
$$x(t)=\frac{\sin(\alpha t)}{\pi t}\tag{4}$$
the integral $(2)$ only converges for $\textrm{Re}\{s\}=0$.
Hence, for the sequence $(3)$ and the function $(4)$, the (discrete-time or continuous-time) Fourier transforms exist. But - as far as I know - one would say that the corresponding $\mathcal{Z}$-transform or Laplace transform does not exist.
My question is: why would it be wrong to claim that in these cases the $\mathcal{Z}$-transform (or Laplace transform) exists with a region of convergence that is reduced to a circle (or a line)? Is it the fact that for the degenerate cases where the region of convergence is reduced to a circle or a line, the corresponding transforms are not analytic? Does it also have to do with the type of convergence of the sum $(1)$ or the integral $(2)$?