Just to give an idea of why the answer should be no: suppose $X$ is a random variable. To make life easy, I'll only talk about even moments, so no cancellation.

Jensen's inequality says $\mathbb{E} X^4\ge(\mathbb{E} X^2)^2$, with equality if and only if $X^2$ is almost surely constant. So not only is this an extra constraint (the second and fourth derivatives of your MGF at zero satisfy an inequality) but also if the constraint holds with equality it tells you all the even moments.

Similarly, $\mathbb{E} X^8\ge (\mathbb{E}X^4)^2$, and similarly equality if and only if $X^2$ is a.s. constant. What if this constraint doesn't actually hold with equality, but it's close to equality? Then $X^2$ has to be quite close to constant, and in particular $\mathbb{E} X^4$ has to be close to $(\mathbb{E} X^2)^2$ (one can quantify this). The higher moments in general tend to control the lower moments. (The other way round is false; putting a bit of probability mass on a very large number doesn't change lower moments much but does push up higher moments).

So even trying to characterise what are possible moments is not eay (and most things are not); and that's simply describing the derivatives at zero of an MGF. Now of course the point of an MGF is that it is its Taylor series at zero (i.e. it's analytic), but if you don't put that in as a condition, I can add an arbitrary function which is smooth, passes through zero and all of whose derivatives at zero are zero (non-constant such functions are well known to exist, e.g. $e^{-1/x^2}$).

aremore conditions. But anyway it is better to look at characteristic functions: the characteristic function of $X_1+X_2$ is the product of characteristic functions of $X_1$ and $X_2$, and if if $1/k$ is not an integer, then the zeroes of these functions mismatch. $\endgroup$ – Mateusz Kwaśnicki Nov 28 at 22:11