How to show that $\text{gcd}(a,m)=d\land \text{gcd}(b,m) = 1 \implies (ab,m) = d$? I was wondering whether my proof would be correct. Thank you in advance!

From Bezout's we have that, $$bx+my=1,\text{ and}$$ $$ax'+my'=d.$$
  Multiplying through by d we get, $$bdx+mdy=d.$$ Substituting we get, $$b(ax'+my')x+mdy=d$$ $$abx'x+bmy'x+mdy=d$$ $$ab(x'x)+m(by'x+dy)=d,$$ as desired.

 A: The “converse” of Bézout doesn't hold except for the case the greatest common divisor is $1$.
What you have proved is just that the greatest common divisor of $ab$ and $m$ is a divisor of $d$.
It’s easy to note, however, that $d$ divides both $ab$ and $m$, so you can conclude. 
A: Consider the  equation 
$$
bx_1+mx_2=d
$$
where $x_1=ax'x$ and $x_2=by'x+dy$, and use a contradiction argument.
A: Bezout's theorem is not an "if and only if" statement unless $\mathrm{gcd}$ is $1$. However, we have the following theorem which is true and you can use it to solve your problem:
$\star$ Theorem: $ax+by=c$ has integer solutions if and only if $(a,b) \mid c$.
Proof: Suppose that $(a,b) \mid c$. Then by Bezout's theorem: $$\exists x,y \in \mathbb{Z}: ax_0 + by_0 = (a,b)$$
Since $(a,b) \mid c$, $\exists q\in \mathbb{Z}: q(a,b)=c$ which implies that $a(x_0q)+b(y_0q)=q(a,b)=c$.
For the other direction, suppose that $ax+by=c$ for some $x,y \in \mathbb{Z}$. We have $(a,b) \mid a$ and $(a,b) \mid b$. Therefore $(a,b) \mid ax+by=c$. $\fbox{ Q.E.D. }$
Back to your proof, your proof can be corrected like this:
$$(b,m)=1 \implies \exists x,y : bx+my=1$$
$$\exists x,y : ab(x)+m(ay)=a \implies (ab,m) \mid a$$
But $$(ab,m) \mid m$$
Therefore, $(ab,m) \mid (a,m)=d$. The other direction is equally easy and I leave it to you. 
A: Bezout's identity is of the form:  
$A\implies B$
Where $A$ is "$\gcd(a,b) = d$" and $B$ is "There are integers $x,y$ so that $xa + yb = d$.
However 
$B \not \implies A$.
As a counter example:
Consider $6$ and $15$, then $x  = 4; y = -1$ then $6x + 15y = 6*4 - 15 = 9$.  But $\gcd(6,15) \ne 9$.
So  $ab(x'x)+m(by'x+dy)=d$ is not at all what was desired.
.....
Instead 
Let $\gcd(ab,m) = e$ then there are $x,y$ so that $abx + ym = e$.  Then as $d|a$ and $d|m$ we must have $d|e$.
If we divide both sides of $abx +ym = e$ we have $\frac aebx + y\frac me = 1$.  And if we multiply both sides by $d$ we have $ ab\frac dex + m\frac dey = d$.  And as $e|ab$ and $e|m$ we must have $e|d$.
So we have $d|e$ and $e|d$ and as we are presuming these are positive integers. That must mean $d = e$.
