Define the identity element.

An element $e \in \mathbb{R}$ is the identity element if for all $a \in \mathbb{R}$, $a \triangle e = a$ and $e \triangle a = a$.
Using $a \triangle e = a$:

Subtract $a$ from both sides:

Factor out $e$:

For this equation to hold for all $a \in \mathbb{R}$, $e$ must be 0. (If $e \neq 0$, then $1 - 2a = 0 \implies a = \frac{1}{2}$, which means $e$ would only work for a specific $a$, not all $a$.)
We can verify with $e \triangle a = a$: $0 \triangle a = 0 + a - 2(0)a = a$, which is correct.
So, the identity element is $e = 0$.

An element $a' \in \mathbb{R}$ is the inverse of $a \in \mathbb{R}$ if $a \triangle a' = e$ and $a' \triangle a = e$.
We found the identity element $e = 0$.
So, we need to solve $a \triangle a' = 0$.

Using the operation $a \triangle a' = a + a' - 2aa'$, we set it equal to 0:

Rearrange the terms to isolate $a'$:

Factor out $a'$ from the left side:

Divide by $(1 - 2a)$ to solve for $a'$:

The inverse element $a'$ exists for all $a \in \mathbb{R}$ such that $2a - 1 \neq 0$, i.e., $a \neq \frac{1}{2}$.

The inverse element is $\boxed{\frac{a}{2a - 1}, \text{ for } a \neq \frac{1}{2}}$.

The inverse of an element $a$ is given by $a' = \frac{a}{2a - 1}$.

To find the inverse of $H$, we replace $a$ with $H$ in the formula:

This inverse exists for $H \in \mathbb{R}$ such that $H \neq \frac{1}{2}$.

The inverse of H is $\boxed{\frac{H}{2H - 1}, \text{ for } H \neq \frac{1}{2}}$.