A simple modular arithmetic query. Given $a,b,c\in\Bbb N$ with $\mathsf{gcd}(a,b)=\mathsf{gcd}(b,c)=\mathsf{gcd}(c,a)=1$ we know that there are $m_1,m_2,m_3\in\Bbb N$ such that $a\equiv m_1a^2\bmod abc$, $ab\equiv m_2ab\bmod abc$ and $b\equiv m_3b^2\bmod abc$ holds.
It is also easy to see there is a single $m$ such that $$a\equiv ma\bmod ab,\quad b\equiv mb\bmod ab$$ holds.
However how to find a single $m$ coprime to $c$ such that $$1\equiv ma\bmod abc,\quad 1\equiv mb\bmod abc$$ holds?
At least how to find a single $m$ such that $$\ell_1 a\equiv ma^2\bmod abc, \quad \ell_2 ab\equiv mab\bmod abc,\quad\ell_3b\equiv mb^2\bmod abc$$ holds where $0<\ell_1,\ell_2,\ell_3<\log abc$ holds and at least one of $\ell_1,\ell_2,\ell_3$ is distinct?
If not how small can we make $\max(\ell_1,\ell_2,\ell_3)$ where  $a\nmid\ell_1$ and $b\nmid\ell_3$ holds?
 A: I'll answer one of your questions ...

You ask how to find a single $m$ such that
\begin{align*}
1 &\equiv ma \text{ mod } (abc)\\
1 &\equiv mb \text{ mod } (abc)\\
\end{align*}
But
\begin{align*}
&1 \equiv ma \text{ mod } (abc)\\
\implies\; &1 \equiv ma \text{ mod } (a)\\
\implies\; &1 \equiv 0 \text{ mod } (a)\\
\implies\; &a = 1\\
\end{align*}
Similarly
\begin{align*}
&1 \equiv mb \text{ mod } (abc)\\
\implies\; &1 \equiv mb \text{ mod } (b)\\
\implies\; &1 \equiv 0 \text{ mod } (b)\\
\implies\; &b = 1\\
\end{align*}
Thus, unless $a = b = 1$, there will be no such $m$.
A: 
At least how to find a single $m$ such that $$\ell_1 a\equiv ma^2\bmod
 abc, \quad \ell_2 ab\equiv mab\bmod abc,\quad\ell_3b\equiv mb^2\bmod
 abc$$ holds where $0<\ell_1,\ell_2,\ell_3<\log abc$ holds?

First note that $gcd(ma^2,abc)=a, gcd(ma^2,a\ell_1)=a$. This is a direct result of $\gcd(a,r)=a \iff \gcd(a,b)=a$ in $a \equiv r \pmod b$
Then rewrite $ma^2 \equiv \ell_1 \pmod{abc}$ as:
$ma^2 = abck + a\ell_1$
$ma = bck + \ell_1$
Then $ma \equiv \ell_1 \pmod{bc}$
It is easy to follow these simplification to obtain the others.

You can simplify as:
$$\ell_1 \equiv ma \pmod{bc}, \quad \ell_2 \equiv m \pmod c, \quad \ell_3 \equiv mb \pmod{ac}$$
Now choose a value $r \in (0,\log abc)$
Note that if $m<c$ then $m=r$ and $\ell_1=\ell_2=\ell_3=r$ 
When $c \mid m$ then $\ell_1=\ell_2=\ell_3=0$
When $m>c \Rightarrow$ $m=ck+r, a=b$ then $\ell_1=\ell_2=\ell_3=r$ and you will have:
$$a\ell_1 \equiv ma \pmod{ac}, \quad \ell_2 \equiv m \pmod c, \quad b\ell_3 \equiv mb \pmod{bc}$$
For $m>c$ then $\ell_1\neq\ell_2\neq\ell_3 \iff a,b,c$ are pairwise coprime thus the solution must follow these restrictions:
$$\ell_1 < bc < \log abc$$
$$\ell_2 < c < \log abc$$
$$\ell_3 < ac < \log abc$$
