Let $A$ be an algebra over the complex numbers (not necessary unital or commutative). An ideal $I$ in $A$ is called modular if there exists a non-zero $e\in A$ such that $ea-a\in I$ and $ae-a\in I$ for all $a\in A$. Equivalently, $e+I$ is a unit for the quotient algebra $A/I$. In this case, we call $e$ a modular element for $I$. Also, it is easy to see that $e\in I$ if and only if $I=A$.
If $A$ is unital, then $1$ is a modular element for every ideal in $A$.
Although I have not seen a counterexample yet, I am quite sure that a modular element is not unique in general. Namely, if $e,f\in A$ are both modular elements for $I$, then we only know that $e+I=f+I$ by uniqueness of the unit and this does not imply that $e=f$.
Now I was wondering if we can make assumptions on $I$ that give uniqueness of $e$.
My first guess was maximality: Can we conclude that $e=f$ when $I$ is maximal?
(If no, are there other conditions on $I$ that guarantee uniqueness?)
EDIT: My second guess is the assumption that we can write $I=\ker(\phi)$ for some non-zero homomorphism $\phi\colon A\to\mathbb{C}$. Then $I$ is automatically a modular maximal ideal (so this case is included in my other guess). I can prove that $\phi(e)=1=\phi(f)$...
It looks like you didn't notice that if $i$ is any element of $I$ at all,
$(e+i)a-a=(ea-a)+ia\in I$ for every $a\in A$, making every element in the coset $e+I$ a modular element for $I$.
That means there's only one condition to make it unique: $I=\{0\}$.