My first attempt to this question was to find the first few moments about the mean and try to rearranging the those moments to obtain the general function as desired. However, when I tried to calculate the first moment about the mean, I got something like
M'(t)=-((e^(t-1))/t^2)+((e^(t-1))/t)
With all those denominator t and t^2, how can find the moments of this function?
The notation $\mu'_r$ refers to the $r^{\rm th}$ raw moment of the random variable $X$; i.e., $$\mu'_r = \operatorname{E}[X^r]$$ whenever this expectation exists. We also have, for some function $g$ of a random variable $X$, $$\operatorname{E}[g(X)] = \int_{x\in \mathcal X} g(x) f_X(x) \, dx,$$ where $\mathcal X$ is the support of $X$, and $f_X$ is the density function of $X$.
So this suggests choosing $g(x) = x^r$, and given $X \sim \operatorname{Uniform}(0,1)$, we know $f_X(x) = 1$ on the support $\mathcal X = [0,1]$, the interval from $0$ to $1$. It follows that $$\operatorname{E}[X^r] = \int_{x=0}^1 x^r \, dx,$$ and the rest is left as an exercise for the reader.
Note that the moment generating function $$M_X(t) = \operatorname{E}[e^{t X}]$$ is not particularly relevant to this question: while it is possible to compute the raw moments this way, it is much more unwieldy: we would need to recall that $$\operatorname{E}[X^r] = \frac{d^r}{dt^r} \left[ M_X(t) \right]_{t=0},$$ and there is no guarantee that the calculation of the $r^{\rm th}$ derivative of the MGF will be tractable.
Nevertheless, it may be illustrative to try it: $$M_X(t) = \int_{x=0}^1 e^{tx} \, dx = \left[\frac{1}{t}e^{tx}\right]_{x=0}^1 = \frac{e^t - 1}{t} = t^{-1}(e^t - 1).$$ By the product rule, $$\frac{dM}{dt} = -t^{-2}(e^t - 1) + t^{-1}e^t,$$ and evaluating at $t = 0$ now becomes a tedious exercise in computing a limit. Indeed, even if we were to obtain a general expression for the $r^{\rm th}$ derivative of $M$, the issue of computing the limit is not going to go away. Clearly, the direct approach is superior.