The standard way to proceed here is by right-tail truncation (see e.g. Section 1 of Nagaev's paper), as follows.

Suppose that $EX_i=0$ for all $i$; the condition that the $X_i$'s are identically distributed will not be used in this answer. Let $S_N:=\sum_{i=1}^N X_i$. Let $T_N:=\sum_{i=1}^N Y_i$, where $Y_i:=X_i\,1(X_i\le b)$ for some real $b>0$. Then
\begin{align}
P(S_N\ge t)&\le P(\max_{i=1}^N X_i>b)+P(T_N\ge t) \\
&\le\sum_{i=1}^N P(X_i>b)+P(T_N\ge t). \tag{0}\label{0}
\end{align}

Finally, to upper-bound $P(T_N\ge t)$ for real $t\ge0$, one can use e.g. Hoeffding's bound (2.9), to get
$$\begin{align}
P(T_N\ge t)
&\le\exp\Big(-\frac tb\,
g\Big(\frac{B^2}{bt}\Big)\Big) \tag{10}\label{10} \\
&\le\Big(\frac{B^2}{bt}\Big)^{ct/b}, \tag{20}\label{20}
\end{align}$$
where $B:=\sqrt{\sum_{i=1}^N EX_i^2}$, $g(u):=(1+u)\ln(1+\frac1u)-1$,
and $c:=\min_{0<u<1}\dfrac{g(u)}{\ln\frac1u}=0.707\ldots$. In particular, choosing $b=ct/p$ for an arbitrary real $p>0$, from \eqref{20} we get
$$P(T_N\ge t)\le\Big(\frac pc \frac{B^2}{t^2}\Big)^p. \tag{30}\label{30}$$

The bound in \eqref{10} is the best bound in its terms based on the Bernstein--Chernoff inequality. Other, similar, more general, and/or better results can be found e.g. in mentioned Section 1 of of Nagaev's paper and in this paper; see also references therein.

Another bound on $P(S_N\ge t)$ for real $t>0$ can be obtained from the Rosenthal inequality -- see e.g. this question and this answer:
$$P(S_N\ge t)\le\frac{E|S_N|^p}{t^p}
\le \frac{K(p)}{t^p}\,(A_p+B^p)$$
for real $p\ge2$, where $K(p)$ is a real number depending only on $p$ and $A_p:=\sum_{i=1}^N E|X_i|^p$. (Actually, the Rosenthal bound can be derived from \eqref{0} and \eqref{30}.)

For $p\in[1,2]$, one can similarly use the von Bahr--Esseen inequality, to get
$$P(S_N\ge t)\le\frac{E|S_N|^p}{t^p}
\le \frac{L(p)}{t^p}\,A_p,$$
where $L(p)\in[1,2]$ depends only on $p$.