Let $a_1/b_1,\ldots,a_n/b_n$ be rational numbers in lowest terms. If $M={\rm lcm}(b_1,\ldots,b_n)$, prove $\gcd(Ma_1/b_1,\ldots,Ma_n/b_n)=1$.

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.

With your added assumption

$$\gcd(a_1, \, \ldots \, , a_n) = 1 \tag{1}\label{eq1A}$$

then the statement of

$$\gcd\left(\frac{Ma_1}{b_1}, \, \ldots \, , \frac{Ma_n}{b_n}\right) = d \tag{2}\label{eq2A}$$

having $d = 1$ is true. To prove this, assume instead that $d \gt 1$ so there's a prime $p \mid d$. There are $2$ basic cases to consider.

First, there's at least one $1 \le i \le n$ where $p \mid \frac{M}{b_i}$. This means there's one or more $1 \le j \le n$ where $p \mid b_j$, so there's also one or more indices where the exponent of $p$ in the prime factorization of $b_j$ is a maximum, call it $m$. Choose any one of these indices, call it $k$. The definition of $\operatorname{lcm}$ means that $M$'s exponent of $p$ in its prime factorization is also the maximum among all of the $b_i$, so it's the same as that of $b_k$, i.e., it's $m$ for both, meaning that $p \not\mid \frac{M}{b_k}$. As such, due to $d$ being the $\gcd$ in \eqref{eq2A}, this requires that $p \mid a_k$ instead. However, since $p \mid b_k$, this means $\frac{a_k}{b_k}$ is not in lowest terms, which is not allowed.

Thus, the second case must apply, i.e., there are no $i$ where $p \mid \frac{M}{b_i}$. This means $p \mid a_i$ for all $1 \le i \le n$ instead. However, this contradicts your added assumption in \eqref{eq1A}.

These contradictions in the $2$ possible cases mean the original assumption of $d \gt 1$ must be false, i.e., it must be that $d = 1$ in \eqref{eq2A} instead.