Here's my attempt to make a proof:
If $[c_1; c_2, \ldots, c_n]$ is any continued fraction, defined in the usual way:
$$ [c_1; c_2, \ldots, c_n] = c_1 + \frac{1}{c_2+\frac{1}{\ldots + \frac{1}{c_n}}}$$
and $\frac{p_m}{q_m}$ is its $m^{th}$ convergent, then, for all $m$ between $1$ and $n$,
$$\frac{p_m}{p_{m-1}} = [c_m;c_{m-1}, \ldots, c_1].$$
This can be proved by induction, using the well-known relation $$p_m = c_m p_{m-1} + p_{m-2} ~~~(1 \leq m \leq n, ~p_0 = 1, ~p_{-1} = 0).$$ The sequential product of the quotients $\frac{p_m}{p_{m-1}}$ is clearly a telescoping product,
$$ p_n = \frac{p_n}{p_{n-1}}\frac{p_{n-1}}{p_{n-2}} \ldots \frac{p_1}{p_0} = \prod_{i=1}^n [c_i;c_{i-1}, \ldots,c_1],$$
where $p_n$ is the numerator of the irreducible fraction equal to $[c_1;c_2,\ldots,c_n]$. Substituting $c_i$ for $a_i$ and $b_i$, we have two products, each equal to the numerator of the respective $n^{th}$ convergent, say $p_n^{(a)}$ and $p_n^{(b)}$. Now, as you already pointed out in your definition of the $b_i$, we have
$$ [b_1;b_2, \ldots, b_n] = [a_n,a_{n-1},\ldots,a_1] = \frac{p_n^{(a)}}{p_{n-1}^{(a)}},$$
which implies $p_n^{(b)} |~ p_n^{(a)}$. However, it's also clear that
$$ [a_1,a_2, \ldots, a_n] = [b_n,b_{n-1},\ldots,b_1] = \frac{p_n^{(b)}}{p_{n-1}^{(b)}},$$
which implies $p_n^{(a)} |~ p_n^{(b)}$. Therefore, $|p_n^{(a)}| = |p_n^{(b)}|=p$, and a more rigorous result follows:
$$ p = \left| \prod_{i=1}^n [a_i;a_{i-1}, \ldots,a_1]\right| = \left| \prod_{i=1}^n [b_i;b_{i-1}, \ldots,b_1]\right|.$$
This is also true for the continued fractions of the kind:
$$ [c_1';c_2',\ldots,c_n'] = c_1 - \frac{1}{c_2-\frac{1}{\ldots - \frac{1}{c_n}}},$$
since $c_i' = (-1)^{i+1}c_i$. The result will be basically the same, only changing the coefficients ($a_i$ for $a_i'$; $b_i$ for $b_i'$) and the values of the convergents ($p_n$ for $p_n'$).
If I didn't get sloppy anywhere, this shows you were pretty right. I hope this could be helpful, since I can't tell much about this particular result.
Let $R = \Bbb Z/N\Bbb Z$. I'll closely follow this 2002 paper by V. P. Elizarov.
Lemma 4.2. If $a, b \in R$ and $\gcd(a,b)=d$ in $\Bbb Z$, then there exists a matrix $Q \in GL_2(R)$ s.t. $(a,b)Q = (d,0)$.
Proof. If $a = 0$ or $b = 0$, take $Q$ to be the identity or an antidiagonal matrix with $1$'s. Suppose $a, b \not = 0$. Over $\Bbb Z$, express $d = a u_1 + b v_1$ by Bezout's identity. If $a = df$ and $b = dg$, then $1 = f u_1 + g v_1$ in $\Bbb Z$.
Furthermore, denote $u = u_1 \mod N$ and $v = v_1 \mod N$ ($u,v \in R$); then we have $d = au+bv$ in $R$, $1=fu+gv$ in $R$. Take $Q = (\begin{smallmatrix}u&g\\v&-f\end{smallmatrix})$ (it seems that the second column in his paper was upside down). Then $AQ = (au + bv, ag-bf) = (1,0)$; the determinant is $-uf-vg=-1$ which means $Q$ is invertible.
Statement 4.3. If $\gcd(a_1, \ldots, a_n) = d$, then there exists $V \in GL_n(R)$ s.t. $AV = (d, 0, \ldots, 0)$, where $A = (a_1,\ldots,a_n)$.
Proof. Start from $(a_{n-1}, a_n)$ and successively eliminate the nonzero elements using Lemma 4.2 (going backwards). Then take the product of these matrices, padded with blocks of identity transformations.
Special case of Statement 4.4. The set of solutions of $A (x_1, \ldots, x_n)^T = 0$ over $R$, where $\gcd(\{a_i\}, n) = 1$, is $V (0, *, \ldots, *)^T$ (the asterisks are arbitrary elements of $R$).
The proof is said to be "a direct check".
Denote $S = (0, *, \ldots, *)^T$. Then, for $A$ and $B$, we can find the corresponding matrices $V_A$ and $V_B$ and ask whether $V_A S = V_B S$, or equivalently, $V_B^{-1}V_A S = S$. Both sides have equal cardinalities, so we only need to check that $V_B^{-1} V_A e_k \in S$ for $k = 2, \ldots, n$ ($\{e_k\}$ is a "standard basis" of $S$, i.e. vectors of all zeros except one $1$).
Best Answer
It is true when $\,d := \gcd\{a_i\} = 1.\,$ More generally it is the special case $\,d = 1,\,m = \ell\,$ of the Theorem below, which is the integral-scaled version of the gcd formula for reduced fractions $\frac{a_i}{b_i}$ (putting $\,m=1\,$ below gives the $k$-ary extension of this formula), which says the the gcd of reduced fractions is the gcd of their numerators over the lcm of their denominators (and dually for lcm).
Theorem $\, $ If $\,m,a_i,b_i\in\Bbb Z,\,$ $\,b_i\mid m a_i,\,$ $\,\color{#c00}{\gcd(b_i,a_i/d)\!=\!1},\,$ $\,d\! =\! \gcd\{a_i\},\,$ $\,\ell\! =\! {\rm lcm}\{b_i\}\,$ then
$\qquad\qquad\quad \gcd\left(\!\dfrac{ma_1}{b_1},\ldots,\dfrac{ma_k}{b_k}\!\right)\, =\, \dfrac{m\gcd(a_1,\ldots,a_k)}{{\rm lcm}(b_1,\:\!\ldots,\:\!b_k)\!\!\!\!\!\!}\,=\,\dfrac{md}{\ell}$
$\begin{align}{\bf Proof}\ \ \ c\,&\mid \gcd\{ma_i/b_i\}\\[.1em] \iff \, \ c\,&\mid {ma_i}/{b_i},\,\forall i, \ \rm by\ gcd\ \color{#90f}{universal}\ property\\[.3em] \iff\! cb_i&\mid ma_i,\, \forall i\\[.3em] \iff\ \color{#c00}{b_i}&\mid ma_i/c={md}/c\,(\color{#c00}{a_i/d}),\, \forall i,\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\smash{\overbrace{\&\ \ md/c\in\Bbb Z}^{\textstyle c\mid ma_i\Rightarrow c\mid\gcd\{ma_i\}= md\!\!\!\!\!\!\!\!\!\!\!}}}\\[.1em] \iff\ b_i&\mid {md}/{c},\,\forall i,\ \ {\rm by\ } \color{#c00}{\gcd(b_i,a_i/d)\!=\!1}\ \text{& Euclid's Lemma}\\[.1em] \iff\ \ell\ &\mid {md}/c,\ \ \ \ \ \ \ \ \rm by\ lcm\ \color{#90f}{universal}\ property\\[.1em] \iff\:\! \ell c&\mid md\\[.1em] \iff\ c\ &\mid {md}/\ell \end{align}$
We used the gcd distributive law in $\,\gcd\{ma_i\} = m\gcd\{a_i\} = md,\,$ and Euclid's Lemma and the gcd & lcm $\rm \color{#90f}{universal}\ properties$.
Remark $ $ Since the proof used only laws valid in every gcd domain, the theorem remains true if we replace $\Bbb Z$ by any gcd domain, e.g. any UFD.