Note that Cramer's theorem applies in dimension 1 w/out any moment assumptions, but of course the rate function may vanish (for this version see theorem 2.2.3 in Dembo-Zeitouni). So the question is equivalent to the question ``when does the rate function vanish'' (the rate function is $\sup_{\lambda\in R}
(\lambda x-\Lambda(\lambda))$, where $\Lambda(\lambda)= \log E(e^{\lambda X_1}))$. So if
$\Lambda(\lambda)=\infty$ for all $\lambda\neq 0$, clearly there is no exponential convergence, not just a failure of the upper bound.
A Remark on the Convolution of the Generalised Logistic Random Variables, Matthew Oladejo Ojo (2003).
There are typos in the article linked above. In Equation (1.1), the denominator of $f$ is $(1+e^x)^{p+q}$. In Section 2, in the unnumbered second equation expressing the caracteristic function of $Y$, the argument of the Gamma function in the denominator is $p$ and not $i p$.
Probability density function
$$ f(x) = \frac{\Gamma(p+q)}{\Gamma(p)\Gamma(q)} \frac{e^{px}}{(1+e^x)^{p+q}}, \quad -\infty<x<\infty,\ p>0,\ q>0$$
for each of $n$ independent random variables gives the probability distribution $f_n(y)$ for the sum:
$$ f_n(x) = e^{px} (\Gamma(p)\Gamma(q))^{-n} \sum_{k=0}^\infty \frac{(-1)^{n(k+1)+1}}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}}\left[ e^{xz}\frac{\Gamma(z+p+q)^n}{\Gamma(z+1)^n} \right]_{\vert z=k}
$$
For example, when $p=q=1$, we get
$$ f_n(x) = e^{x} \sum_{k=0}^\infty \frac{(-1)^{n(k+1)+1}}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}}\left[ e^{xz} (z+1)^n \right]_{\vert z=k}
$$
Some explicit expressions of $f_n(y)$ for this case $p=q=1$:
$$f_{1}(y)=\frac{e^y}{\left(e^y+1\right)^2}$$
$$f_{2}(y)=\frac{e^y \left(e^y (y-2)+y+2\right)}{\left(e^y-1\right)^3}$$
$$f_{3}(y)=\frac{e^{2 y} \left[-2 y^2+\left(y^2+6\right) \cosh y-6 y \sinh y+6\right]}{\left(e^y+1\right)^4}$$
$$f_{4}(y)=\tfrac{1}{3}e^{5y/2}(e^y-1)^{-5}[y (11 y^2-36) \cosh(y/2) + y (36 + y^2) \cos(3 y/2) -
24 (2 y^2-2 + (2 + y^2) \cosh y) \sinh(y/2)]$$
Best Answer
The term "sub-Gaussian" usually refers to random variables with tail decaying like that of a Gaussian, not exponential.
Now for a random variable $X$ being sub-Gaussian in this sense is equivalent to any of the following conditions:
A similar set of equivalent conditions characterize random variables with sub-exponential tails:
For an explanation of that, see, e.g, the first chapter of Lugosi's lecture notes on measure concentration.
Now in your case the sub-Gaussian tail estimate for $X_i$ is easy to check. From there you get an $O(p^{1/2})$ moment estimate on them, therefore an $O(p)$ moment estimate on the product, therefore an exponential tail estimate on the product.