Proving Closure Under Multiplication in a Set with Specific Properties

I encountered a problem where I needed to prove that a set ( A ) is closed under multiplication, given the following properties:

1. $0 \in A$
2. $1 \in A$
3. If $x, y \in A$, then $x - y \in A$
4. If $x \neq 0$, then $\frac{1}{x} \in A$
5. If $x, y \in A$, then $x +y \in A$

To prove that $xy \in A$ for $x, y \in A$, I followed these steps:

1. If $x \neq 0$ and $x \neq 1$, since $x - 1 \in A$ and $x \in A$:$ \frac{1}{x-1} \in A \quad \text{and} \quad \frac{1}{x} \in A$

   Therefore:

   $\frac{1}{x-1} - \frac{1}{x} = \frac{x - (x-1)}{x(x-1)} = \frac{1}{x(x-1)} \in A$

   Consequently:

$ x(x-1) \in A \quad \Rightarrow \quad x^2 - x \in A \quad \Rightarrow \quad x^2 \in A $

2. Similarly for $y$:$ y^2 \in A $

3. Since $x + y \in A$, we have:$ (x + y)^2 \in A$

   Thus:

$ (x + y)^2 - (x^2 + y^2) = 2xy \in A $

4. Therefore:$ \frac{1}{2xy} \in A$

   This implies:

$ \frac{1}{2xy}+\frac{1}{2xy} = \frac{1}{xy} \in A \quad \Rightarrow \quad xy \in A $

My Questions:

1. Why does this algebraic manipulation prove that $xy \in A$?
2. Is there an alternative logical approach to prove the closure under multiplication for set $A$?

Any additional insights or corrections to my approach would be greatly appreciated.

Thank you!