Let me abstract a little bit the situation and show that the inequality you have in mind stems from a rather general result (as mentioned in the comments, the following should be explained in every introductory text on large deviations principles). Consider some integrable random variables $\xi$ and $(\xi_n)_{n\geqslant1}$ and, for some $n\geqslant1$ and some positive $x$, the event
$$
A_n=[\xi_1+\cdots+\xi_n\geqslant n\mathrm E(\xi)+nx].
$$
Using first $\xi_i=\log\phi(X_i)$ and then $\xi_i=-\log\phi(X_i)$, the event you consider is the union of two events $A_n$, hence our task is to bound $\mathrm P(A_n)$.
Now comes an additional, non trivial, hypothesis: assume that $\xi$ is exponentially integrable, that is:
$\qquad\qquad$ There exists $t_0\gt0$ such that $\mathrm E(\mathrm e^{t\xi})$ is finite for every $|t|\lt t_0$. $\qquad(\ast)$
Then the proof is simple. Start from the almost sure inequality, valid for every nonnegative $t$:
$$
\mathrm e^{nt(\mathrm E(\xi)+x)}\,\mathbf 1_{A_n}\leqslant\mathrm e^{t(\xi_1+\cdots+\xi_n)}=\mathrm e^{t\xi_1}\cdots\mathrm e^{t\xi_n}.
$$
Integrate both sides and use the independence assumption. This yields
$$
\mathrm e^{nt(\mathrm E(\xi)+x)}\,\mathrm P(A_n)\leqslant\mathrm E(\mathrm e^{t\xi})^n.
$$
We proved that, for every nonnegative $t$,
$$
\mathrm P(A_n)\leqslant u(t)^n\qquad\text{where}\qquad u(t)=\mathrm E(\mathrm e^{t\xi})\mathrm e^{-t(\mathrm E(\xi)+x)}.
$$
Note that a lot of these inequalities are just useless. For example, $u(0)=1$ hence the $t=0$ version is $\mathrm P(A_n)\leqslant 1$... Big deal! Even worse: if $t\gt0$ is such that $u(t)\gt1$, or such that $\mathrm E(\mathrm e^{t\xi})$ is infinite, we painfully proved that $\mathrm P(A_n)$ is bounded by something greater than $1$, or infinite. :-)
On the other hand, if there exists at least some nonnegative $t$ such that $u(t)\lt1$, the inequality becomes a strong control on $\mathrm P(A_n)$, namely, that $\mathrm P(A_n)\to0$ at least as fast as a geometric sequence. Hence the key question becomes:
$\qquad\qquad\qquad$ Does there exist some $t\gt0$ such that $u(t)\lt1$?
To answer this, let us consider the limit $t\to0$ with $t\gt0$. Then,
$$
\mathrm E(\mathrm e^{t\xi})=1+t\mathrm E(\xi)+o(t),\qquad
\mathrm e^{-t(\mathrm E(\xi)+x)}=1-t(\mathrm E(\xi)+x)+o(t),
$$
hence $u(t)=1-tx+o(t)$. Since $x\gt0$, this limited expansion implies that $u(t)\lt1$ for (at least) some (small) $t\gt0$. The proof is over.
To sum up, the result stems from a Markov-type inequality plus a limited expansion of the Laplace transform $t\mapsto\mathrm E(\mathrm e^{t\xi})$ at $t=0$. The Markov-type inequality is powerful because the $\xi_n$ are independent. The limited expansion is valid because we assumed $(\ast)$.
Note that the proof above uses the finiteness of $\mathrm E(\mathrm e^{t\xi})$ for $0\leqslant t\leqslant t_0$ only. The finiteness for $-t_0\leqslant t\leqslant 0$ is needed when dealing with the event $[\xi_1+\cdots+\xi_n\leqslant n\mathrm E(\xi)-nx]$, hence the control $(\ast)$ must indeed be two-sided.
Coming back at last to the question you asked, if $\xi=\log\phi(X)$ with $X$ centered normal of variance $\sigma^2$ and $\phi$ the density of the distribution of $X$, then $\mathrm E(\mathrm e^{t\xi})$ is finite if and only if $\mathrm E(\phi(X)^{t})$ is finite if and only if $\mathrm E(\mathrm e^{-tX^2/(2\sigma^2)})$ is finite if and only if $t+1\gt0$, hence $(\ast)$ holds with $t_0=1$.
