Well, this is not an exact answer, but we can use the following thoughts to prove the statement without using Rolle's Theorem.
Since it is given that $f' \left( a \right) = f' \left( b \right) = 0$ and assuming that $f$ is a non - constant polynomial (otherwise, if it is a constant non - zero poynomial, then it has no roots and hence the statement is true), we can see that either $f \left( a \right) \geq 0$ or $f \left( a \right) \leq 0$. Without loss of generality, I will give the arguments for $f \left( a \right) \geq 0$.
If $f \left( a \right) = 0$, clearly, $a$ is a root. Now, since $f$ is a polynomial which is non constant, there is a neighbourhood of $a$ where $f$ is either increasing or decreasing. We shall deal with the case where $f$ is increasing (similar argument can be given for decreasing).
Now, suppose that the neighbourhood is smaller than $\left( a, b \right)$ and after that neighbourhood, the function again starts decreasing (or becomes constant). Then, $\exists c \in \left( a, b \right)$ such that $f' \left( c \right) = 0$, contradicting the hypothesis. Therefore, the neighbourhood is atleast as much as $\left( a, b \right)$. Since $f$ is increasing in this interval, and $f \left( a \right) = 0$, there cannot be any other root in this interval. Hence, $f$ has at most one root.
Now, the other case where $f \left( a \right) > 0$. Again, from the above arguments, $f$ is either increasing or decreasing in $f \left( a, b \right)$. Hence, if it is increasing, $f \left( b \right) > f \left( a \right) > 0$ and it has no root in $\left( a, b \right)$. Similarly, if it is completely decreasing, it can cross the $x -$ axis atmost once.
Therefore, in all the cases, we can say that $f$ has atmost one root.