This can be proved through repeated application of the triangle inequality.
Let the difference $D_k = a_k\cdots a_n-b_k\cdots b_n$, so
Note that
$$
\begin{align}
D_1 & = a_1a_2\cdots a_n-b_1b_2\cdots b_n \\
& = (a_1 - b_1)b_2\cdots b_n + a_1(a_2\cdots a_n -b_2\cdots b_n) \\
& = (a_1 - b_1)b_2\cdots b_n + a_1 D_2
\end{align}
$$
which allows for a recursion, i.e. an equivalent formula can be constructed again for $D_2$ etc.
For the absolute values, we use the triangle inequality:
$$
\begin{align}
|D_1| & = | (a_1 - b_1)b_2\cdots b_n + a_1 D_2| \\
& \leq | a_1 - b_1 | |b_2\cdots b_n| + |a_1| | D_2|
\end{align}
$$
with equality if and only if the phase of $(a_1 - b_1)b_2\cdots b_n$ equals the phase of $a_1 D_2$. Repeated use results in
$$
D_1 = \sum_{k= 1}^n c_k ( a_k - b_k)
$$
with $c_k = a_1 \cdots a_{k-1} \cdot b_{k+1} \cdots b_n$. Hence
$$
|D_1| \leq \sum_{k= 1}^n |c_k | \; | a_k - b_k|
$$
Since all $|a_1|, |a_2|, .., |a_n|, |b_1|, |b_2|, ..., |b_n| \in (0, 1]$, we have all $|c_k | \leq 1$, and therefore
$$
|D_1| \leq \sum_{k= 1}^n | a_k - b_k|
$$
with equality if and only if for all terms with $a_k - b_k \neq 0$, the $|c_k| = 1$ and the phases of the corresponding $c_k (a_k - b_k) $ are equal.
Since the term $a_1 - b_1 \neq 0$, $c_1 (a_1 - b_1)= (a_1 - b_1) b_2\cdots b_n$ sets the phase and $|c_1| =1 $ requires that for all $k \geq 2$, all $|b_k| =1 $.
Now suppose also $a_2 - b_2 \neq 0$. Then $c_2 = a_1 b_3\cdots b_n$ and $|c_2| =1 $ is only possible if $|a_1| =1 $. So if $|a_1| < 1 $, we have a contradiction. Hence for $|a_1| < 1 $, the OP's desired equation is only possible when $a_2 - b_2 = 0$. Continuing with the following terms, the same argument gives that for all $k \geq 2$, $a_k - b_k = 0$. By exchange of $a_k$ and $b_k$, the same argument holds for $|b_1| < 1 $.
This proves the OP's claim
for $|a_1| < 1 $ or $|b_1| < 1 $.
$\quad \quad \Box$
P.S.: For $|a_1| = |b_1| = 1 $, I am still lacking an argument for the proof.
