In the most layman terms it's a way to encode all characteristics of the probability distribution into one short phrase. For instance, if I know that MGF of the distribution is $$M(t)=e^{t\mu+1/2\sigma^2t^2}$$
I can find out the mean of this distribution by taking first term of Taylor expansion:
$$\left. \frac d {dt}M(t) \right|_{t=0}=\mu+\sigma^2t \Bigg|_{t=0}=\mu$$
If you know what you're doing it's much faster than taking the expectation of the probability function. 

Moreover, since this MGF encodes *everything* about the distribution, if you know how to manipulate the function, you can apply operations on all characteristics of the distribution at once! Why don't we always use MGF? First, it's not in every situation the MGF is the easiest tool. Second, MGF doesn't always exist.

# Above layman
Suppose you have a standard normal distribution. You can express everything you know about it by stating its PDF: $$f(x)=\frac 1 {\sqrt{2\pi}}e^{-x^2/2}$$

You can calculate its moment such as mean and standard deviation, and use it on transformed variables and functions on random normals etc.

You can think of the MGF of normal distribution as an alternative to PDF. It contains the same amount of information. I already showed how to obtain the mean. 

Why do we need an alternative way? As I wrote, sometimes it's just more convenient. For instance, try calculating the variance of the standard normal from PDF:
$$\sigma^2=\int_{-\infty}^\infty x^2\frac 1 {\sqrt{2\pi}}e^{-x^2/2} \, dx=\text{?}$$
It's not that difficult, but it's much easier to do it with MGF $M(t)=e^{t^2/2}$:
$$\sigma^2= \left. \frac {d^2} {dt^2}M(t) \right|_{t=0} = \left. \frac d {dt} t \, \right|_{t=0}=1$$