$\newcommand{\proj}[1]{{{\mathrm{proj}}(#1)}}$
$\newcommand{\ideal}[1]{{\mathfrak #1}}$
Let $\phi:S \to T$ be an isomorphism $\phi_d:S_d \to T_d$ for all $d \geqslant d_0$. Then $\proj{S} \cong \proj{T}$ and the question is: Given a homogeneous prime $\ideal{p}$ of $S$, what is the corresponding prime $\ideal{q}$ of $T$, so that $\phi^{-1}(\ideal{q}) = \ideal{p}$.
I claim, that
$$\ideal{q} = (\sqrt{\phi_{\geqslant d_0}(\ideal{p}_{\geqslant d_0}) T_+}: b)$$
where $b \in T_d$ with $d > d_0$ but not in $\sqrt{\phi_{\geqslant d_0}(\ideal{p}_{\geqslant d_0}) T_+}$.
First it is clear, that $\ideal{p}_{\geqslant d_0}$ is a homogeneous ideal of $S$. Next it follows, that $\phi(\ideal{p}_{\geqslant d_0}) T T_+$ is an homogeneous ideal of $T$. It is identical with $\phi(\ideal{p}_{\geqslant d_0}) T_+$. Call that ideal $\ideal{q}''$ and call $\ideal{q}' = \sqrt{\ideal{q}''}$.
Now for $b_1 b_2 \in \ideal{q}'$ and $b_1,b_2 \in T_+$ it follows that $(b_1 b_2)^n \in \ideal{q}''$. Therefore $(b_1 b_2)^{n n'} \in \phi(\ideal{p}_{\geqslant d_0})$ and $b_i^{n n'} \in T_{e_i}$ with $e_i > d_0$, therefore $b_i^{n n'} = \phi(a_i)$ with $a_i \in S_{e_i}$. So we have $\phi(a_1)\phi(a_2) = \phi(a)$ with $a \in \ideal{p}_{e}$ with $e > d_0$. So $a_1 a_2 = a$ and without restriction of generality $a_1 \in \ideal{p}$. So $b_1^{n n'} = \phi(a_1) \in \phi(\ideal{p}_{\geqslant d_0})$ and so $b_1^{n n'} \in \ideal{q''}$, therefore $b_1 \in \sqrt{\ideal{q}''} = \ideal{q}'$.
So the homogeneous ideal $\ideal{q}'$ fulfills a weak primality for $b_1,b_2 \in T_+$. From this follows the strong primality of $\ideal{q} = (\ideal{q}':b)$. For let $b_1 b_2 \in \ideal{q}$ therefore $b_1 b_2 b \in \ideal{q}'$ ($b_1, b_2$ homogeneous in $T$ of arbitrary degree) then $(b_1 b) (b_2 b) \in \ideal{q}'$. Therefore because $b b_i \in T_+$ without restriction of generality $b_1 b \in \ideal{q}'$, that is $b_1 \in \ideal{q}$.
The last thing is to prove $\phi^{-1}(\ideal{q}) = \ideal{p}$. Let $\phi(a) \in \ideal{q}$ ($a$ homogeneous in $S$) that is $\phi(a) b \in \ideal{q'}$. Then $(\phi(a) b)^n \in \ideal{q}''$ and $(\phi(a)b)^{n n'} \in \phi(\ideal{p}_{\geqslant d_0})$. Now $b^{n n'} = \phi(a_1)$ with $a_1 \notin \ideal{p}$ and the right side is $\phi(a_2)$ with $a_2 \in \ideal{p}$. So we have $a^{n n'} a_1 = a_2 \in \ideal{p}$ therefore $a^{n n'} \in \ideal{p}$, therefore $a \in \ideal{p}$ as was to be shown.
P.S. I used in the above, that for $b \in \ideal{q}'' = \phi(\ideal{p}_{\geqslant d_0}) T_+$, $b$ homogeneous, a high power $b^N$ is in $\phi(\ideal{p}_{\geqslant d_0})$. This is obvious from the equation
$$b = b_1 z_1 + \cdots + b_r z_r$$
with $b_i \in \phi(\ideal{p}_{\geqslant d_0})$ and $z_i \in T_+$.
7.4.M. : Let $Z$ be a constructible subset of $X$ and $\pi: X\to Y$.
Assume that image of any finite type morphism of noetheian schemes in constructibe. Now $Z$ is a disjoint union of $Z_i$ where each $Z_i$ is a locally closed subset of $X$. So to show $\pi(Z)$ is constructible, it suffices to show each $\pi(Z_i)$ is constructible. Now $Z_i=U_i\cap F_i$ where $U_i$ is an open subset and $F_i$ is a closed subset of $X$. Put the reduced induced structure on $F_i$, that induces a scheme structure on the open subset $Z_i=U_i\cap F_i\subset F_i$. Now consider the map $$\phi|_{Z_i}:Z_i\to Y $$ given by the restriciton of the map $$\phi: F_i\to X\to Y.$$
Now by our assumption image of $\phi|_{Z_i}=\pi(Z_i)$ is constructible. So this proves. 7.4.M.
7.4.N. : (First reduction) Now we want to reduce to the case $Y$ is affine. We have $\pi:X\to Y$. Now cover $Y$ by affine open subsets $Spec ~A$'s. Now $$\pi(X)=\bigcup_i \pi( \pi^{-1}(Spec~ A_i)).$$
So it suffices to show image of $\pi: \pi^{-1}(Spec~ A_i)\to Spec ~ A_i$ is constructible.
So we have reduced to the case where $Y$ is affine.
(second reduction):
Now we want to reduce to the case $X$ is affine. We have $\pi:X\to Y$ with $Y$ affine.
Again cover $X$ be affine open $Spec ~B$'s. Now $$\pi(X)=\bigcup_i\pi( Spec~B_i).$$ So it suffices to consider $X$ affine.
I hope this clears up the sequence of reductions.
Best Answer
Just to get this off the unanswered list, your solution is totally correct and does not need the finite generation hypothesis (see for instance here). As Misha Lavrov points out in the comments, Vakil has only made his definition of $S_{n\bullet}$ for finitely generated $S_\bullet$, which is why the question is stated as it is.