First, a few simple definitions:
Definition: $A$ is called a $*$-algebra if it is equipped with a map $x \mapsto x^*$ satisfying the following: (i) $1^* =1$, (ii) $(xy)^* = y^*x^*$ for every $x,y \in A$, and (iii) $(ax+y)^* = \overline{a} x^* + y^*$ for every $x,y \in A$ and $a \in \Bbb{C}$.
Definition: Let $A_h = \{a \in A \mid a^* = a \}$ be the set of all hermitian/self-adjoint elements. The $*$-positive cone, denoted by $A_+$, is the collection of all hermitian elements satisfying: (i) $\Bbb{R}_{\ge 0} 1 \subseteq A_+$; (ii) $ax +y \in A_+$ for all $x,y \in A_+$ and $a \in \Bbb{R}_{\ge 0}$; and (iii) $x^* A_+ x \subseteq A_+$ for every $x \in A$.
One can quickly see from the above definition that $A_+$ contains elements of the form $x^*x$ (and sums of such elements), as expected. Moreover, given the above definition, we can say that for $x,y \in A_h$, $x \le y$ means $y-x \in A_+$.
Definition: $A_b = \{x \in A \mid x^*x \le M1 \mbox{ for some } M > 0)\}$ is the set of all bounded elements
Definition: A unital $*$-algebra $A$ is called a semi-pre-$C^*$-algebra provided $A = A_b$ and it comes with a $*$-positive cone.
My question is, for any semi-pre-$C^*$-algebra $A$, why is
$$I(A) := \{x \in A \mid x^*x \le \epsilon 1 \mbox{ for all } \epsilon > 0\}$$ in fact an ideal? I was able to show closure under the involution, under left and right multiplication, but was not able to show closure under addition. If $x,y \in I(A)$, then $(x+y)^*(x+y) = x^*x + x^*y + y^*x + y^*y$. Clearly $x^*x$ and $y^*y$ can be bound above by any multiple of $\epsilon$, but how does one deal with $x^*y + y^*x$, which is roughly the real part of $x^*y$?
Note that $0\leq(x-y)^*(x-y)=x^*x-x^*y-y^*x+y^*y$, so $x^*y+y^*x\leq x^*x+y^*y$.