If $(m_1,m_2)=D$, $(a,m_1)=d_1$, $(b,m_2)=d_2$, then $(am_2+bm_1, m_1m_2)=?$ We know, If $(m_1,m_2)=(a,m_1)=(b,m_2)=1, \iff (am_2+bm_1, m_1m_2)=1$
I tried to generalize now.
Let $(a,m_1)=d_1, (b,m_2)=d_2$ where $d_1,d_2$ need not to be 1. 
$(am_2+bm_1, m_1m_2)$
$=(am_2+bm_1, m_1)(am_2+bm_1, m_2)\ as\ (m_1,m_2)=1$
$=(am_2, m_1)(bm_1, m_2)$
$=(a, m_1)(b, m_2)\ as\ (m_1,m_2)=1$
So, $(am_2+bm_1, m_1m_2)=(a, m_1)(b, m_2)$
Now, I want to make $(m_1,m_2)=D$ where D is not necessarily 1.
Let $\frac{a}{A}=\frac{m_1}{M_1}=d_1$  and $\frac{b}{B}=\frac{m_2}{M_2}=d_2$,
so $(A,M_1)=1$ and $(B,M_2)=1$
$(am_2+bm_1, m_1m_2)=d_1.d_2(AM_2+BM_1, M_1M_2)=(a,m_1)(b,m_2)(AM_2+BM_1, M_1M_2)$
Now let $\frac{M_1}{M_{11}}=\frac{M_2}{M_{22}}=D_{12}$ i.e., $(M_1,M_2)=D_{12}$
so, $(M_{11},M_{22})=1$
then $(am_2+bm_1, m_1m_2)$
$=(a,m_1)(b,m_2)D_{12}(AM_{22}+BM_{11}, M_{11}M_{22}D_{12})$
$=(a,m_1)(b,m_2)(M_1,M_2)(AM_{22}+BM_{11}, M_{11}M_{22}D_{12})$
But, this $D_{12}$ is not necessarily co-prime with  $M_{11}$ or $M_{22}$.
So, I could not proceed any further.
 A: Part 1:
We have both
$$
(am_2+bm_1,m_1m_2)=d_1d_2\,\left(\frac{a}{d_1}\!\!\frac{m_2}{d_2}+\frac{b}{d_2}\!\!\frac{m_1}{d_1} ,\frac{m_1}{d_1}\!\!\frac{m_2}{d_2}\right)\tag{1}
$$
and
$$
(am_2+bm_1,m_1m_2)=D\,\left(a\frac{m_2}{D}+b\frac{m_1}{D} ,m_1\frac{m_2}{D}\right)\tag{2}
$$
Equations $(1)$ and $(2)$ show that
$$
\mathrm{lcm}(d_1d_2,D)\,\vert\,(am_2+bm_1,m_1m_2)\tag{3}
$$

Part 2:
Since $(m_1,m_2)=D$, let
$$
m_1x+m_2y=D\tag{4}
$$
Since $(a,m_1)=d_1$, let
$$
au_1+m_1v_1=d_1\tag{5}
$$
Since $(b,m_2)=d_2$, let
$$
bu_2+m_2v_2=d_2\tag{6}
$$
Now, $(4)$ and $(5)$ yield
$$
\begin{align}
(am_2+bm_1)y+m_1(ax-by)&=aD\\
(am_2+bm_1)u_1+m_1(m_2v_1-bu_1)&=m_2d_1
\end{align}\tag{7}
$$
and $(4)$ and $(6)$ yield
$$
\begin{align}
(am_2+bm_1)x+m_2(by-ax)&=bD\\
(am_2+bm_1)u_2+m_2(m_1v_2-au_2)&=m_1d_2
\end{align}\tag{8}
$$
Using $(7)$ and Bezout, we can write
$$
(am_2+bm_1)w_1+m_1z_1=(aD,m_2d_1)\tag{9}
$$
Using $(8)$ and Bezout, we can write
$$
(am_2+bm_1)w_2+m_2z_2=(bD,m_1d_2)\tag{10}
$$
Thus, taking the product of $(9)$ and $(10)$ yields
$$
(am_2+bm_1)w+m_1m_2z_1z_2=(aD,m_2d_1)(bD,m_1d_2)\tag{11}
$$
and therefore, $(11)$ and Bezout yield
$$
(am_2+bm_1,m_1m_2)\,\vert\,(aD,m_2d_1)\,(bD,m_1d_2)=d_1d_2D^2\,\left(\frac{a}{d_1},\frac{m_2}{D}\right)\,\left(\frac{b}{d_2},\frac{m_1}{D}\right)\tag{12}
$$

Conclusion:
The best I have come up with so far is
$$
\mathrm{lcm}(d_1d_2,D)\,\vert\,(am_2+bm_1,m_1m_2)\,\vert\,d_1d_2D^2\,\left(\frac{a}{d_1},\frac{m_2}{D}\right)\,\left(\frac{b}{d_2},\frac{m_1}{D}\right)\tag{13}
$$
