The condition that $Z \to X$ is of finite presentation means that $Z$ is cut out by a finite type quasi-coherent sheaf of ideals $\mathcal{I} \subset \mathcal{O}_X$, see Morphisms, Lemma \ref{morphisms-lemma-closed-immersion-finite-presentation}. Write $\mathcal{A} = \bigoplus_{n \geq 0} \mathcal{I}^n$ so that $X' = \underline{\text{Proj}}(\mathcal{A})$. Note that $X \setminus Z$ is a quasi-compact open of $X$ by Properties, Lemma \ref{properties-lemma-quasi-coherent-finite-type-ideals}. Since $b^{-1}(X \setminus Z) \to X \setminus Z$ is an isomorphism (Lemma \ref{lemma-blowing-up-gives-effective-Cartier-divisor}) the same result shows that $b^{-1}(X \setminus Z) \setminus Z'$ is quasi-compact open in $X'$. Hence $U = X \setminus (Z \cup b(Z'))$ is quasi-compact open in $X$. By Lemma \ref{lemma-closed-subscheme-proj-finite-type} there exist a $d > 0$ and a finite type $\mathcal{O}_X$-submodule $\mathcal{F} \subset \mathcal{I}^d$ such that $Z' = \underline{\text{Proj}}(\mathcal{A}/\mathcal{F}\mathcal{A})$ and such that the support of $\mathcal{I}^d/\mathcal{F}$ is contained in $X \setminus U$. \medskip\noindent Since $\mathcal{F} \subset \mathcal{I}^d$ is an $\mathcal{O}_X$-submodule we may think of $\mathcal{F} \subset \mathcal{I}^d \subset \mathcal{O}_X$ as a finite type quasi-coherent sheaf of ideals on $X$. Let's denote this $\mathcal{J} \subset \mathcal{O}_X$ to prevent confusion. Since $\mathcal{I}^d / \mathcal{J}$ and $\mathcal{O}/\mathcal{I}^d$ are supported on $X \setminus U$ we see that $V(\mathcal{J})$ is contained in $X \setminus U$. Conversely, as $\mathcal{J} \subset \mathcal{I}^d$ we see that $Z \subset V(\mathcal{J})$. Over $X \setminus Z \cong X' \setminus b^{-1}(Z)$ the sheaf of ideals $\mathcal{J}$ cuts out $Z'$ (see displayed formula below). Hence $V(\mathcal{J})$ equals $Z \cup b(Z')$. It follows that also $V(\mathcal{I}\mathcal{J}) = Z \cup b(Z')$ set theoretically. Moreover, $\mathcal{I}\mathcal{J}$ is an ideal of finite type as a product of two such. We claim that $X'' \to X$ is isomorphic to the blowing up of $X$ in $\mathcal{I}\mathcal{J}$ which finishes the proof of the lemma by setting $Y = V(\mathcal{I}\mathcal{J})$. \medskip\noindent First, recall that the blowup of $X$ in $\mathcal{I}\mathcal{J}$ is the same as the blowup of $X'$ in $b^{-1}\mathcal{J} \mathcal{O}_{X'}$, see Lemma \ref{lemma-blowing-up-two-ideals}. Hence it suffices to show that the blowup of $X'$ in $b^{-1}\mathcal{J} \mathcal{O}_{X'}$ agrees with the blowup of $X'$ in $Z'$. We will show that $$ b^{-1}\mathcal{J} \mathcal{O}_{X'} = \mathcal{I}_E^d \mathcal{I}_{Z'} $$ as ideal sheaves on $X''$. This will prove what we want as $\mathcal{I}_E^d$ cuts out the effective Cartier divisor $dE$ and we can use Lemmas \ref{lemma-blow-up-effective-Cartier-divisor} and \ref{lemma-blowing-up-two-ideals}. \medskip\noindent To see the displayed equality of the ideals we may work locally. With notation $A$, $I$, $a \in I$ as in Lemma \ref{lemma-blowing-up-affine} we see that $\mathcal{F}$ corresponds to an $R$-submodule $M \subset I^d$ mapping isomorphically to an ideal $J \subset R$. The condition $Z' = \underline{\text{Proj}}(\mathcal{A}/\mathcal{F}\mathcal{A})$ means that $Z' \cap \Spec(A[\frac{I}{a}])$ is cut out by the ideal generated by the elements $m/a^d$, $m \in M$. Say the element $m \in M$ corresponds to the function $f \in J$. Then in the affine blowup algebra $A' = A[\frac{I}{a}]$ we see that $f = (a^dm)/a^d = a^d (m/a^d)$. Thus the equality holds.