$P\{X_1+...+X_n\geq n\epsilon\} \le \prod_{j=1}^n P\{X_j \geq\epsilon\}$ since there are also other ways that the sum can be greater than or equal to $n\epsilon$. So now you want to show that $P\{X_j\ge\epsilon\}\le M(t)e^{-t\epsilon}$.
This is actually known as exponential Chebyshev inequality. To derive this, note by the same reasoning that $P\{X\ge\epsilon\}\le P\{|X|\ge\epsilon\}$ (I'm not sure whether this simplification is really necessary). Then we need a general version of the Chebyshev inequality thanks to @sos440 (see her/his comment below), which is also (up to absolute value) equivalent to its general measure-theoretic statement given on Wikipedia. The simplification step seems to be necessary if using @sos440's derivation, but not if one uses the general measure-theoretic statement from Wikipedia which does not require the absolute value.
The existence of a finite bound $K$ actually guarantees that $X$ has finite moments of each order, so that $E[e^{tX}]$ is finite.
Note, however, that Chebyshev's inequality -- and Bernstein's inequalities, for that matter -- are derived from the more basic Markov's inequality, $P\{|X|>a\}\le\frac{E[|X|]}{a}$. In particular, the latter uses $E[e^{t\sum X_j}]$ in its derivation, so your original thought of using $E[e^{tX_j}]$ might also bear fruit by applying Markov's inequality if you wanted to try your hand at a custom derivation.