I've got a problem from a STEP III paper which I am trying to answer. 
I have got as far as
$$M_T (\theta) = \frac{1}{(1-\theta)^2}$$
and
$$ \int_{0}^{\infty} g(u) e^{\theta u} \mathrm{d}u = \frac{1}{(1-\theta)^4}$$
which I need to solve for the probability density function $g(u)$. I do remember a bit about moment generating functions, how to obtain them, their independence and whatnot, but I am struggling to figure out how to show that the density function is that as described in the question. As this is a STEP paper (and some knowledge of moment generating functions is needed) is there a way of doing this without using Fourier/Laplace transforms as this isn't in the A-level curriculum? I am quite out of practice and am figuring there must be a much simpler way to do this (i.e. perhaps integrating by parts and then setting up a differential equation in $g(u)$)?
I'm not sure what's a valid method here by your standards, but one way to figure this out is just pattern recognition.
You computed the MGF for the PDF $f(t)=te^{-t},$ presumably like this: $$ M(\theta) = \int_0^\infty e^{\theta t}te^{-t}dt = \int_0^\infty te^{-(1-\theta)t}dt = \frac{1}{(1-\theta)^2}\int_0^\infty ue^{-u}du = \frac{1}{(1-\theta)^2}.$$ Notice this is the square of the MGF for an exponential distribution $e^{-t}$, and it's easy to see why: we have that extra $t$ out front that results in an extra factor of $(1-\theta)^{-1}$ when we do the substitution.
From this, it is natural to consider the family of distributions $f(t) = \frac{1}{n!}t^{n}e^{-t}$ for which we have, similarly to before, $$ M(\theta) = \int_0^\infty e^{\theta t}\frac{1}{n!}t^{n}e^{-t}dt = \frac{1}{(1-\theta)^{n+1}} \frac{1}{n!}\int_0^\infty u^{n}e^{-u}du = \frac{1}{(1-\theta)^{n+1}}.$$ So your desired distribution is for $n=3,$ which corresponds to the pdf $f(t) = \frac{1}{3!}t^3e^{-t}.$