Describing the set ${nin Bbb Nlvert (n>1)wedge (forall x,yin Bbb N)[(xy=n)implies (x=1lor y=1)]}$

Let's break it down step by step.

1. You're first point is right, for $n in A$, $n$ must be an natural number greater than $1$.

2. $n$ must also satify $(xy=n)implies(x=1vee y=1)$ for all natural numbers $x$ and $y$, which can be $1$.

3. Now let's look at the general cases for the form of $Aimplies B$

If $A$ is false, then $(Aimplies B)$ is true. If $A$ is true, then $(Aimplies B)$ is true if an only if $B$ is also true.

So now we know that if

$(xy=n)$ is false for all natural numbers of $x,y$, then $nin A$. However, it's very easy to see that it's not always false, because if $x=1$ and $y=n$ (which is a perfectly valid possibility!), then $(xy=n)$ is clearly true.

What that means is that for the cases where $(xy=n)$ is true, we must also have that $(x=1vee y=1)$ to be true. That is the part that restricts $n$ to be a prime number.

I.e; if $n$ is a multiple of two natural numbers $x,y$, it's only prime when for all $x,y$, either $x$ or $y$ is $1$.

Your third point is false. An implication is not necessarily falsified when the conclusion is false.

For instance, (False) $implies$ (False) is always true.

It might be appropriate to recall the definition : $A implies B$ holds if, whenever $A$ holds, $B$ holds as well. If $A$ is false, then the implication $A implies B$ is true, by definition.