You should probably state what you want a little more precisely. As it is currently stated, it allows for the possibility that $\pi$ is not projective in which case there is no chance. The statement also allows a small resolution in which case $E$ is empty and $a\pi^*L$ is not ample for any $a$.

Unfortunately, even if you assume that $\pi$ is projective and $E$ is the entire exceptional set(!), it is possible that this fails. The condition you need is that $-E$ has to be relatively ample for $\pi$. This can fail already for surfaces. For instance, assume that $E$ has two components, both with negative self-intersection, say $-n$ and $-m$, and let's say that the intersection number of the two components is $n+e$ for some positive number $e$. The intersection matrix has to be negative definite, which means that we need $nm> (n+e)^2$, but this is easy to satisfy by making $m$ really big. In this case, $E$ restricted to the component with self-intersection $-n$ has positive degree, so $-E$ cannot be ample. 

This suggests that you cannot allow $E$ to have more than one component. On the other hand, in that case you are OK. Alternatively, if you allow different coefficients for the components of $E$ then a similar statement holds.

**Addendum:** In response to @freidtchy's comment-question below here is an explanation of the last sentence above. 
Yes, I meant exactly that there exist $a_i>0$ such that $A\pi^*L-\sum a_iE_i$ is ample. Since $\pi$ is ample, there exists a $\pi$-ample Cartier divisor on $Y$. With a little bit of work one may assume that there is one which is entirely supported on $E$ (the possible components that are a priori not can be exchanged to something pulled back from $X$ plus something supported on $E$ and the pull-back of anything is $\pi$-trivial). In other words, one has a $\pi$-ample Cartier divisor $\sum b_iE_i$. Then the Negativity Lemma [Kollár-Mori-98, 3.39] tells us that all the $b_i$ are negative and then choosing a large enough $A$ gives the claimed statement.