Update: As Eric Wofsey mentioned in the comments to his answer, there is a 3rd condition which is equivalent to (1) and (2):
(3) If $f = \sum_{d \ge 0} f_d \in I$, where for each $d$, $f_d \in R_d$, then $f_d \in I$ for each $d \ge 0$.
Since $\mathbb{C}[x_0, \dots, x_n] = \oplus_{d \ge 0} R_d$, every $f \in \mathbb{C}[x_0, \dots, x_n]$ is of the form $f = \sum_{d \ge 0} f_d$ where for each $d \ge 0$, $f_d \in R_d$. So (3) is equivalent to:
(3') If $f \in I$, then $f_d \in I$ for all $d \ge 0$.
Which is itself equivalent to:
(1') $I \subseteq \oplus_{d \ge 0} (I \cap R_d)$. (Since $f_d \in R_d$ by definition, so if it is also in $I$, then $f_d \in I \cap R_d $.)
(Skipping fewer steps, (3) is equivalent to the statement:
If $f \in I \cap ( +_{d \ge 0} R_d)$, then $f \in +_{d \ge 0} (I \cap R_d)$. I.e. intersection with $I$ distributes over the sum.
But anyway, we know that $+_{d \ge 0} R_d = \oplus_{d \ge 0} R_d = \mathbb{C}[x_0, \dots, x_n]$, so $I \cap (+_{d \ge 0} R_d) = I \cap (\oplus_{d \ge 0} R_d) = I \cap \mathbb{C}[x_0, \dots, x_n] = I$.
Moreover, $+_{d \ge 0} (I \cap R_d) = \oplus_{d \ge 0}(I \cap R_d)$, which follows from the direct sum decomposition of $\mathbb{C}[x_0, \dots, x_n]$, because $(\cap_{d \not= e} R_d) \cap R_e = \{0\} \implies I \cap [(\cap_{d \not= e} R_d) \cap R_e ] = \{0\} \iff [I \cap ( \cap_{d \not= e} R_d)] \cap [I \cap R_d] = \{0 \}$.
So [If $f \in I \cap ( +_{d \ge 0} R_d)$, then $f \in +_{d \ge 0} (I \cap R_d)$] $\iff$ [If $f \in I$, then $f_d \in I$ for all $d \ge 0$.])
Anyway the moral of this story is: it is easier to prove $\neg(2) \implies \neg (3)$.
First, let's state correctly $\neg(2)$, as Eric Wofsey explained how to do in his answer:
$\neg(2)$ Every set of generators for $I$ contains a non-homogeneous polynomial.
Let's state $\neg (3)$ (or something equivalent to it):
$\neg(3)$ There is some element of $I$ which has a homogeneous part which is not in $I$.
This is clearly true in the case that $I$ is the principal ideal generated by some non-homogeneous polynomial $f$, since no homogeneous polynomial is in $I$, so $f$ has (at least two) homogeneous parts which are not in $I$ (because the homogeneous parts are homogeneous polynomials, and no homogeneous polynomial is in $I$).
So proving $\neg(2) \implies \neg(3)$ would clearly generalize the argument from the special case that $I$ is a principal ideal generated by a non-homogeneous polynomial, which is what I wanted to do.
So let's show more generally that $\neg(2) \implies \neg (3)$. Assume that every generating set of $I$ has at least one non-homogeneous polynomial in it. Then assume by means of contradiction that (3) were true (that $I = \oplus_{d \ge 0} (I \cap R_d)$). Then $\cup_{d \ge 0}(I \cap R_d)$ would be a generating set for $I$ consisting exclusively of homogeneous polynomials. (Because any generating set for $I$ as an abelian group also has to be a generating set for $I$ as an ideal, since $I$ is by assumption closed under $\mathbb{C}[x_0, \dots, x_n]$ multiplication.) This would be a contradiction, since $\neg(2)$ says that every generating set of $I$ contains at least one non-homogeneous polynomial. Therefore (3) $\iff$ (1) is false, i.e. we have $\neg(3) \iff \neg(1)$. (This argument also seems to explain what Eric Wofsey meant in his comment when he said "I don't see a way to prove this is equivalent to $\neg(2)$ without basically going through (1)".
In particular, we would have at least one generating set with a non-homogeneous polynomial with at least one homogeneous part which is not generated by the homogeneous polynomials in $I$ -- this fact generalizes the example of $I = \langle x - yz, x^2 - yz \rangle$ which I gave earlier: $x \not\in I$, and since $x$ is the part of $x- yz \in I$ in $R_1$, we have that $x \not\in I \implies x-yz \not\in \oplus_{d \ge 0}(I \cap R_d) \implies \neg(3)$.
/Update
Attempt:
Because $\mathbb{C}$ is Noetherian, by Hilbert's Basis theorem, so is $\mathbb{C}[x_0, \dots, x_n]$. So (2) is equivalent to:
(2') $I$ is generated by a finite set of homogeneous polynomials.
Also, we have for all $d \ge 0$ that $(I \cap R_d) \subseteq I$ (trivially), and since $I$ is closed under addition, we must have as a result that $\oplus_{d \ge 0} (I \cap R_d) \subseteq I$. Therefore (1) is equivalent to:
(1') $I \subseteq \oplus_{d \ge 0} (I \cap R_d)$.
(I showed that the sum $+_{d \ge 0}(I \cap R_d)$ being a direct sum follows from the fact that the sum $\oplus_{d \ge 0} R_d = \mathbb{C}[x_0, \dots, x_n]$ is direct, which I also previously showed to be true.)
I showed that, for polynomials with field coefficients (more generally coefficients in any null-divisor-free ring) that the product $gf$ of any non-zero polynomial $g$ with a non-homogeneous polynomial $f$ is necessarily non-homogeneous. Thus, for any $d \ge 0$, $\langle f \rangle \cap R_d = \langle 0 \rangle$, with the result that $\oplus_{d \ge 0} (\langle f \rangle \cap R_d) = \langle 0 \rangle$. Clearly $\langle f \rangle \not\subseteq \langle 0 \rangle$, so (1') is false, and thus (1).
In the special case of $\neg (2)$ that $I = \langle x-yz, x^2 - yz \rangle = \langle yz, x^2 -x \rangle$, I have from the special case above that $\langle x^2 - x\rangle \cap R_d = \langle 0 \rangle$, but I'm not sure how to use this to show that $x^2 - x \not\in \oplus_{d \ge 0}(\langle yz, x^2 - x \rangle \cap R_d)$, since non-homogeneous polynomials are not closed under addition. I really want to say that, for all $d \ge 0$, $\langle yz, x^2 - x \rangle \cap R_d = \langle yz \rangle \cap R_d$, which would give me the result I want to show (since $yz$ doesn't divide $x^2 -x$), but (a) I'm not sure how to correctly show that, since non-homogeneous polynomials aren't closed under addition, and (b) even if I did show that, it wouldn't seem to generalize to all possible cases of $\neg (2)$, so it wouldn't be very useful line of thinking to pursue to show that, in general, $\neg (2) \implies \neg (1)$.
For example, if $f_1 = x - y^2$ and $f_2 = y^2 + z$, then $f_1 + f_2$ is homogeneous and non-zero even though both $f_1, f_2$ are non-zero and non-homogeneous, and clearly $f_1 + f_2 \in \langle f_1, f_2 \rangle$, so even though for all $d \ge 0$, $\langle f_1 \rangle \cap R_d = \langle 0 \rangle$ and $\langle f_2 \rangle \cap R_d = \langle 0 \rangle$, we do not have for all $d \ge 0$ that $\langle f_1, f_2 \rangle \cap R_d = \langle 0 \rangle$. Of course, in this case we also clearly have that $\langle f_1, f_2 \rangle = \langle x, z, y^2 \rangle$, which is generated by homogeneous polynomials, so this isn't actually a special case of $\neg (2)$. But even if $\neg (2)$ does exclude all problematic cases like this, I don't know how to state $\neg(2)$ correctly.
Showing $(1) \implies (2)$ directly might be easier, but the only way of proving it I have seen (see here or here) doesn't directly give a finite generating set for the ideal $I$. I would prefer a condition involving finite generating sets, to fully take advantage of the Noetherian assumption.
