Your map $f$ is not well-defined (when $\phi$ is not surjective), since it may happen that $\phi^{-1}(\mathfrak{q})$ contains the irrelevant ideal. Instead of defining $f$ globally, you can use the affine covering of the Proj scheme. It suffices to prove that a) the ring homomorphism $S_{(f)} \to T_{(\phi(f))}$ induced by $\phi$ is an isomorphism for every homogeneous $f \in S$ of positive degree, b) the basic open subsets $D(\phi(f))$ cover $\mathrm{Proj}(T)$.
Hint for b): If $g \in T$ is homogeneous of positive degree, choose some $n \geq 1$ such that $g^n$ has degree $\geq d_0$.
Hint for a): This is easy when you have already digested the idea of b).
$\newcommand{\proj}[1]{{{\mathrm{proj}}(#1)}}$
$\newcommand{\ideal}[1]{{\mathfrak #1}}$
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 $z_i \in T_+$.
Best Answer
Since by definition $\mathfrak{p}$ does not contain all of $T_+ = \oplus_{d>0} T_d$, there is some $s \in T_d$ for some $d>0$, such that $s \notin \mathfrak{p}$. You proved that $\mathfrak{p} \cap T_d = \mathfrak{q} \cap T_d$, so $s \notin \mathfrak{q}$ either.
Let $t \in \mathfrak{p} \cap T_0$. Then $t s \in \mathfrak{p} \cap T_{d} = \mathfrak{q} \cap T_{d}$, hence $t \in \mathfrak{q}$ or $s \in \mathfrak{q}$, and the latter is impossible.