It is the element $-1$.
On a more serious note, the definition of "generates" includes allowing the inverse of the generating elements. For any group $G$, and element $g\in G$, the subgroup generated by $g$ is
$$\{g^n:n\in\mathbb{Z}\}$$
not
$$\{g^n:n\in\mathbb{N}\}$$
(the latter is not a subgroup unless $g$ has finite order).
Observe that $g^{-1}$ is always in $\{g^n:n\in\mathbb{Z}\}$.
You're trying to PROVE that $G$ is cyclic, so you cannot (yet) assert that $x = g^n$. Instead, consider $x \cdot x$. It must be either $x, y,$ or $e$. If it's $x$, then you have
$$
x^2 = x\\
x^2 (x^{-1}) = x x^{-1}\\
x = e
$$
which is a contradiction, because $x$ and $e$ are distinct elements of the group.
If $x^2 = e$, then $x$ has order 2, but 2 does not divide 3, so this contradicts Lagrange's theorem.
Finally, we conclude that $x^2 = y$, and thus the group is cyclic, generated by the element $g = x$.
{Alternative if you don't like Lagrange yet:}
In the case where we suppose that $x^2 = e$:
The elements $xe, xx,$ and $xy$ must all be distinct for if two were the same, then multiplying by $x^{-1}$ on the left would show that two of $e, x, y$ were the same, which is impossible.
Since $xe = x$ and we're assuming $x^2 = e$, we must have
$$
xy = y.
$$
multiplying on the right by $y^{-1}$ gives $x = e$, a contradiction. So $x^2 = e$ is also impossible.
Best Answer
Suppose $\mathbb{R}^*$ is cyclic. Let $a$ be its generator. Since $-1 \in \mathbb{R}^*$, there exists a nonzero integer $n$ such that $-1 = a^n$. Then $a^{2n} = 1$. Hence the order of $a$ is finite. This is a contradiction.