I want to show that the moment generating function is finite everywhere iff $$\lim_{t\rightarrow\infty} e^{\lambda t}P(\lvert X\rvert > t) = 0$$ If we assume the mgf is finite everywhere, then I have the following: $$P(\lvert X\rvert >t) \le P(\lvert X\rvert \ge t) \le \frac{E(e^{\lambda \lvert X\rvert})}{e^{\lambda t}} = \frac{E(e^{\lambda X})\rvert_{X\ge 0} + E(e^{-\lambda X})\rvert_{X<0}}{e^{\lambda t}}\xrightarrow{t\rightarrow\infty} 0$$ This is not sufficient however, as the limit $e^{\lambda t}P(\lvert X\rvert >t)$ might well be constant. Can this approach be fixed somehow or is it flawed?
I don't have any ideas for the reverse direction, how could it be done?
You have shown that if MGF is finite everywhere then $e^{\lambda t}P(|X|\ge t)\to 0$ as $t\to\infty.$ To show the converse, first make some simplification. Replace $X$ with $|X|$ (check that moment generating function of $X$ exists if and only the moment generating function of $|X|$ does so). Now note that $$E\left(e^{\lambda |X|}\right)=\int e^{\lambda x}d\mu_{|X|}(x)=\lambda \int_{0}^{\infty}e^{\lambda t} P(e^{\lambda |X|}\ge e^{\lambda t})dt.$$
The second equality above is a consequence of 'Layer cake representation'. Now use that $P(e^{\lambda |X|}\ge e^{\lambda t})=P(|X|\ge t).$ Now conclude that $e^{\lambda t}P(|X|\ge t)\to 0$ as $t\to \infty.$ (Note that it is easy to conclude that the limit if it exists, has to be zero. The point is to argue that the limit does exist.)