Non-associative commutative "group" When dealing with some hash functions that I was trying to speed up, I toyed with a binary operation with the goal to "approximate" the addition on $\{0,1\}^*$ when seen as binary representation of the positive integers:
$$(a,b) \in (\{0,1\}^*)^2 \mapsto a \oplus b \oplus ((a \land b) \ll 1)$$ where $\oplus$ is bitwise XOR and $\land$ is bitwise AND, and $\ll$ is left-shift by 1 position.  (The purpose of $((a \land b) \ll 1)$ is to simulate the "carry-bit" operation.)
Formal definition. Let $\{0,1\}^\mathbb{N}$ denote the collection of functions $f:\mathbb{N}\to \{0,1\}$ and let $$\{0,1\}^* = \{x \in \{0,1\}^\mathbb{N}: \exists N\in\mathbb{N}(\forall k\in\mathbb{N}(k\geq N\implies x(k)=0))\}.$$
Denote by $\ll 1$ the shift by one position, i.e. $\ll 1 : x \in \{0,1\}^* \to x'\in \{0,1\}^*$ where $x'(0) = 0$ and $x'(n+1) = x(n)$ for all $n\in \mathbb{N}$. We usually write $x \ll 1$ instead of $\ll 1(x)$.
For any $a,b\in\{0,1\}^*$ let us write $$a +_2 b := a \oplus b \oplus ((a\land b) \ll 1).$$
It is easily seen that $+_2$ is not associative, and that $0$ is a neutral element for every $x\in \{0,1\}^*$. Moreover, $+_2$ is clearly commutative
Question. Given $a,b\in \{0,1\}^*$, is there $x\in \{0,1\}^*$ such that $a +_2 x = b$? Is $x$ necessarily unique?
Further question. (Need not be answered for acceptance.) If the answer to the above question is positive, we would have a kind of "non-associative group". Do these have a proper name? Do they occur "naturally" somewhere in mathematics?
 A: The answer to the first question is yes:

Claim. Let $a, b \in \{0, 1 \}^{\ast}$ and let $n \ge 0$ be the least integer such that $a(i) = b(i) = 0$ for every $i > n$. Then the equation $$a +_2 x = b$$ has a unique solution $x \in \{0, 1 \}^{\ast}$ which is recursively defined by $x(0) = a(0) + b(0)$, $x(i) = a(i - 1)x(i - 1) + a(i) + b(i)$ for $0 < i \le n$ and
$x(n + 1) = a(n)x(n)$, $x(i) = 0$ for $i > n + 1$.
Equivalently, the solution $x$ is given by
$$
\begin{array}{lll}
x(0) &=& a(0) + b(0), \\
x(1) &=& a(0)(a(0) + b(0)) + a(1) + b(1),\\
... & = & ..., \\
x(n) & = & a(0) \cdots a(n - 1)(a(0) + b(0)) + a(1) \cdots a(n - 1)(a(1) + b(1)) + \cdots + a(n - 1)(a(n - 1) + b(n - 1)) + a(n) + b(n),
\end{array}$$
$x(n + 1) = a(n)x(n)$
and $x(i) = 0$ for $n > n + 1$ where the operations $+$ and $\cdot$ refer to addition and multiplication in $\mathbb{Z} / 2 \mathbb{Z}$ which we identified with $\{0, 1 \}$.


Proof. The equation is equivalent to $x \oplus ((a \wedge x) \ll 1) = a \oplus b$ which reads as a linear system over $\mathbb{Z} / 2 \mathbb{Z}$.

