What is the motivation of defining the mmoment generating function of a random variable $X$ as: $E[e^{tX}]$? I know that one can obtain the mean, second moment, etc.. after computing it. But what was the intuition in using $e^{x}$? Is it because its onetoone and always increasing?
Take the definition of "generating function" for a sequence. Do it where the sequence is the sequence consisting of the moments of $X$. That's it. 


As you suggest, the fact that $e^x$ is increasing is a useful property here. In particular, that lets you in some cases to apply Markov's/Chebychev's inequality to the random variable $e^{tX}$ in order to get exponentially decaying bounds on the tails of $X$; see e.g. Chernoff bounds. In principle any other positive increasing function could be used in the same way, but $e^x$ is a particularly useful choice because it is especially well suited to studying sums of independent random variables, as noted already in Yuval's answer. 


The goal is to to put all the moments in one package. Since $$ e^{tx} = \sum \frac{x^n}{n!} t^n $$ the coefficients of $t^n$ in $E(e^{tx})$ are (scaled) moments. In other contexts we can use $$ (1xt)^{1} = \sum x^n t^n $$ in place of $e^{tx}$. This gives more or less what engineers call the "ztransform" and in combinatorics it is known as "ordinary generating function". Using the exponential has the happy advantage that convolution of random variables translates to product of moment generating functions. 


If X and Y are independent then $E[e^{t(X+Y)}] = E[e^{tX}] E[e^{tY}]$, so convolution corresponds to multiplication of the mgf's. Another reason: the moment generating function is actually a Fourier transform. Now suppose $X_i$ are i.i.d. with zero mean, and define $Y_n = \sum_{i=1}^n X_i/\sqrt{n}$. Define $\phi(t) = E[e^{tX_1}]$. Then $E[e^{tY_n}] = \phi(t/\sqrt{n})^n$. Under reasonable assumptions, $\phi(t) = 1 + V[X_1]t^2/2 + O(t^3)$, and so $E[e^{tY_n}] = (1 + V[X_1]t^2/2n + O(t^3)/n^{1.5})^n \longrightarrow e^{V[X_1]t^2/2}$, and we get the central limit theorem (by continuity of the Fourier transform). 

