Moment Generating Function for Sum of Independent Random Variables

You have not made use of the definition of a moment generating function. The moment generating function for any random variable $X$ is usually defined as $$M_X(t) = \mathbb{E} \left( e^{tX} \right)$$

EDIT

Adding more details.

First, your $f_y$ is incorrect. It should be $f_Y(y) = \lambda e^{- \lambda y}$. We get $$M_Y(t) = \displaystyle \int_0^{\infty} e^{ty} \lambda e^{-\lambda y} dy = \displaystyle \int_0^{\infty} \lambda e^{(t-\lambda) y} dy = \dfrac{\lambda}{\lambda-t}$$ Hence, the moment generating function for $Y = Y_1 + Y_2 + Y_3$ is $M_Y(t) = \dfrac{\lambda_1}{\lambda_1-t} \dfrac{\lambda_2}{\lambda_2-t} \dfrac{\lambda_3}{\lambda_3-t}$

You might want to check your pdf of $Y$ for typographical errors. Is it a valid pdf?

First, if $Y = aX+b$, then $$f_Y(y) = \frac{1}{|a|}f_X\left(\frac{y-b}{a}\right)$$ and so if $f_X(x) = xe^{-x}$ for $x \geq 0$ $b = 0$, $a = \lambda^{-1}$, then $f_Y(y) = \lambda (\lambda y) e^{-\lambda y}$ for $y \geq 0$. (This is the correct pdf, not what you have, but for an alternative, see the answer by Marvis). Thus, if you can figure out $M_X(t)$, the MGF of $X$, then you can deduce the MGF of $Y$ as $M_Y(t) = M_X(\lambda^{-1}t)$, from which it follows that $M_W(t) = [M_X(\lambda^{-1}t)]^3$.

So what is $M_X(t)$? As Marvis points out to you,

$$M_X(t) = E[e^{tX}] = \int_0^\infty e^{tx} x e^{-x}\,\mathrm dx = \int_0^\infty x e^{(t-1)x}\,\mathrm dx$$ which integral you should be able to compute via integration by parts.