First approach (Evan Chen) We introduce the following notations:
$$\begin{aligned} P(x) &= c_n x^n + c_{n-1} x^{n-1} + \dots + c_1 x + c_0 \\ &= c_n(x + \alpha_1) \dots (x + \alpha_n) \\ P(x) - 1 &= c_n(x + \beta_1) \dots (x + \beta_n) \end{aligned}$$
By taking conjugates,
$$\implies \left(x + \frac{1}{\alpha_1}\right) \cdots \left(x + \frac{1}{\alpha_n}\right) = \left(x + \frac{1}{\beta_1}\right) \cdots \left(x + \frac{1}{\beta_n}\right) + \left(\overline{c_n}\right)^{-1} \quad (\spadesuit)$$
The equation (♠) is the main player:
Claim — We have $c_k = 0$ for all $k = 1, \dots, n-1$.
*Proof.* By comparing coefficients of $x^k$ in (♠) we obtain
$$\frac{c_{n-k}}{\prod_i \alpha_i} = \frac{c_{n-k}}{\prod_i \beta_i}$$
but $\prod_i \alpha_i - \prod_i \beta_i = \frac{1}{c_n} \neq 0$. Hence $c_k = 0$. $\square$
It follows that $P(x)$ must be of the form $P(x) = \lambda x^n - \mu$, so that $P(x) = \lambda x^n - (\mu+1)$. This requires $|\mu| = |\mu+1| = |\lambda|$ which is equivalent to the stated part.

Second approach (from the author) We let $A = P$ and $B = P - 1$ to make the notation more symmetric. We will as before show that $A$ and $B$ have all coefficients equal to zero other than the leading and constant coefficient; the finish is the same.
First, we rule out double roots.
Claim — Neither $A$ nor $B$ have double roots.
*Proof.* Suppose that $b$ is a double root of $B$. By differentiating, we obtain $A' = B'$, so $A'(b) = 0$. However, by Gauss-Lucas, this forces $A(b) = 0$, contradiction. $\square$

Claim (Main claim) — For any $i$ and $j$, $\frac{a_i}{a_j}$ is a power of $\omega$.
*Proof.* Note that
$$\frac{a_i - b_1}{a_j - b_1} \cdots \frac{a_i - b_n}{a_j - b_n} = \frac{B(a_i)}{B(a_j)} = \frac{A(a_i) - 1}{A(a_j) - 1} = \frac{0 - 1}{0 - 1} = 1.$$
Since the points $A_i$, $A_j$, $B_k$ all lie on the unit circle, interpreting the left-hand side geometrically gives
$$\angle A_i B_1 A_j + \cdots + \angle A_i B_n A_j = 0 \implies n \widehat{A_i A_j} = 0,$$
where angles are directed modulo $180^\circ$ and arcs are directed modulo $360^\circ$. This implies that $\frac{a_i}{a_j}$ is a power of $\omega$. $\square$
Now the finish is easy: since $a_1, \dots, a_n$ are all different, they must be $a_1\omega^0, \dots, a_1\omega^{n-1}$ in some order; this shows that $A$ is a multiple of $x^n - a_1^n$, as needed.