I want to understand how this sentence works, as to me it seems 'if and only if' changes meanings of the statements:
Suppose that $a,b\in \mathbb N$. Then there exists a unique $d \in\mathbb N $ for which the following holds:
an integer $m$ is a multiple of $d$ if and only if $m = ax+ by$ for some $x, y\in\mathbb Z$.
"$m = ax + by$ for some integers $x,y.$"
As $m$ is not defined, $m$ can be every number made up that way, which means 'for all'.
On the other if an integer $m$ is a multiple of $d,$ then $m = ax+ by$ for some $x, y \in \mathbb Z,$ which means 'for some' rather than 'for all'.
This means: $$\forall a{\in}\mathbb N\;\forall b{\in}\mathbb N\;\exists! d{\in}\mathbb N\;\forall m{\in}\mathbb N\;\big(\exists k{\in}\mathbb Z\; m=kd\iff \exists x{\in}\mathbb Z\; \exists y{\in}\mathbb Z\; m=ax+by\big).\tag1$$
Well yes, there is an implicit "for all $m$" after the colon.
Statement $(1)$ can be rewritten as $$\forall a{\in}\mathbb N\;\forall b{\in}\mathbb N\;\exists! d{\in}\mathbb N\;\bigg(\forall m{\in}\mathbb N\;\color{red}{\forall k{\in}\mathbb Z}\; \big(m=kd\implies \exists x{\in}\mathbb Z\; \exists y{\in}\mathbb Z\; m=ax+by\big)\\\land \forall m{\in}\mathbb N\;\color{red}{\forall x{\in}\mathbb Z\; \forall y{\in}\mathbb Z}\; \big(m=ax+by\implies \exists k{\in}\mathbb Z\; m=kd\big)\bigg).$$ Does this resolve your query?
Not per se.