This is not true. Actually, your question needs to be made a bit more precise. I suppose you mean that $E$ is the exceptional set and that $\mathscr O_Z(-E)$ denotes the ideal sheaf of $E$. (Notice that $E$ is not necessarily a divisor and then its ideal is not an invertible sheaf). However, even if you assume that $Z$ is smooth, $E$ is a smooth divisor, the statement is still not true. You can even assume that $Y$ is smooth and $X$ is normal.
1
If $X$ is not normal, then $\pi_*\mathscr I_{E\subseteq Z}$ is not even necessarily in $\mathscr O_X$, let alone being equal to the ideal sheaf of $Y$. Just take $X$ to be a cuspidal (or nodal) cubic and $Y$ the singular point. The blow up is the map induced by $$k[t^2,t^3]\hookrightarrow k[t]$$ and $\pi_*\mathscr I_{E\subseteq Z}$ corresponds to the $k[t^2,t^3]$ module $k[t] \cdot t$.
2
So, let's assume that $X$ is normal. Unfortunately it is still not true: Let $X$ be a quadric cone in $\mathbb A^3$ and $Y$ a line through the vertex of $X$. Then the blow up of $X$ along $Y$ is the same as the blow up of the vertex and $\pi_*\mathscr I_{E\subseteq Z}$ is the ideal sheaf of the vertex, not of the line.
Notice that in this example, $X$ is a normal Gorenstein variety and $Y$ is smooth, so you need quite a bit of assumptions to have a blanket statement like you wished for.
3
Here is a criterion that implies what you want:
If $X$ is normal and the natural map $\mathscr O_Y\to \pi_*\mathscr O_E$ is an isomorphism, then $\mathscr I_{Y\subseteq X}\simeq \pi_*\mathscr I_{E\subseteq Z}$. The proof of this is very simple. Consider the diagram
\begin{gather}
0 \quad \longrightarrow & \quad \mathscr I_{Y\subseteq X} \qquad \longrightarrow & \mathscr O_X \ \qquad \longrightarrow &\mathscr O_Y &\longrightarrow & 0\\
&\downarrow \qquad & \downarrow \qquad\qquad & \downarrow & \qquad \qquad \\
0\quad \longrightarrow & {\pi_*\mathscr I_{E\subseteq Z}} \quad \longrightarrow &\pi_* \mathscr O_Z \qquad \longrightarrow &\pi_*\mathscr O_E &
\end{gather}
The assumption that $X$ is normal implies that the second vertical arrow is an isomorphism and the other assumption is that so is the third. Then it follows easily that so is the first.