Consider
$(x+y).(x+y) = x+y$
$\implies (x+y).x + (x+y).y = x+y$
$\implies x.x + y.x + x.y + y.y = x+y$
$\implies x + y.x + x.y + y = x + y$
$\implies x+ y + y.x + x.y = x+ y$
$\implies x.y = - y.x \to (1)$
Now, we have $a.a = a \\ \implies (a+a) . (a+a) = a+a \\ \implies a.a + a.a + a.a + a.a = a + a \\ \implies a.a + a.a = 0 \\ \implies a + a = 0 \\\implies a = -a \to (2)$
From (1) and (2),
$x.y = y.x, \forall x, y$
So, the given ring is commutative.