Let $\mathbb C^{\ast} := \mathbb C \setminus \{0\}$.
So $\mathbb C^{\ast}$ is a group closed under multiplication.
Consider $\langle i\rangle$ which generates $i$.
So,
$i$, $i^2$, $i^3$, $i^4$, $...$, $i^n$ are generated.
Which is the equivalent to:
$i$, $-1$, $-i$, $1$, $...$, $i^n$ are generated.
$1$ is the multiplicative identity.
My guess is that since $\langle i\rangle$ generates identity, that means we can generate the entire group. So the order is infinite.
First, let me take a moment to address your confusions.
That would be the right reasoning for an ideal $I$ in a ring $R$; if $1\in I$, then $I = R$, because an ideal must satisfy $rx\in I$ for any $r\in R$ if $x\in I$. And if $1\in I$, then $r\cdot 1 = r\in I$ for any $r\in R$, implying that $R = I$.
However, we are talking about groups (not rings), and groups have only one identity, as they have only one binary operation (rings have two, addition and multiplication). Any subgroup of $G$ generated by any set of elements must contain the (only) identity of $G$ (as any subgroup contains the identity), and containing the identity does not mean it contains the entire group. For example, consider any group $G$, and the subgroup generated by the identity $e$. Well, $\{e\}$ is already a subgroup, so the subgroup generated by $e$ is just $\{e\}$ ($\langle e\rangle = \{e\}$)!
A group generated by a single element $g$ is necessarily cyclic, since $$ \langle g\rangle = \{g^n\mid n\in\Bbb Z\}, $$ and this admits a surjective map from $\Bbb Z$ given by \begin{align*} \Bbb Z&\to \langle g\rangle\\ n&\mapsto g^n. \end{align*} Thus, $\langle g\rangle$ is a quotient of $\Bbb Z$, and is hence cyclic of some order. (Note that this implies that $\langle i\rangle$ cannot be isomorphic to $\Bbb C^\times$, as $\Bbb C^\times$ is not cyclic.)
Back to the problem at hand. We want to determine what $\langle i\rangle\subseteq\Bbb C^\times$ is. Now, we observe that $i^4 = 1$, which is the identity. So, $i^0 = 1$, $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, $i^5 = i$, and so on (similarly for negative powers of $i$). So the set $$\langle i\rangle = \{i^n\mid n\in\Bbb Z\} = \{\dots, i^{-2}, i^{-1}, i^0, i^1, i^2,i^3, i^4,\dots\} = \{\dots, -1,-i,1, i, -1,-i, 1,\dots\}$$ is really simply the set $\{1, i, i^2, i^3\}$. This tells us that $\#\langle i\rangle = 4$ (the order of $i$), and in particular, that $\langle i\rangle $ is isomorphic to the cyclic group of order $4$.
Why do you believe that $1 \in \langle i \rangle$ implies $i$ generates all of $\mathbb C^*$? Multiplying any number $z$ by $1$ produces $z$ again, so you do not obtain any new numbers in this way.
So what is the order of $\langle i \rangle$?
Since $i^4 = 1$, $$i^n = i^ni^4 = i^{n+4}$$ for all integers $n$. This means that if you listed all the powers of $i$, the values in the list would repeat in cycles of 4. So there are only four distinct powers of $i$, namely $i, -1, -i$, and $1$.