Possessive Kleene star operator
I will share below my attempt to prove that you can express every regular language as a regular expression using just the $+$ and $@$ operators. Unfortunately, my attempt at a proof has a gap.  Perhaps you can repair the gap.  My attempt based upon Brzozoswki's method for generating regular expressions, so I'll assume you are already familiar with how that works.



First, some definitions and other preliminaries.

Definition. Define $A^+$ to be the language of one or more words from $A$, i.e., $A^+ = \{a_1 a_2 \cdots a_k \mid k \ge 1, a_1,\dots,a_k \in A\}$.  In other words, $A^+ = A^* \setminus \{\epsilon\}$.

Definition. $\newcommand{\rp}{\backslash\backslash}$  For sets $A,B \subseteq \Sigma^*$, define $A\rp B$ by $A\rp B = (A^*\backslash B) - (A^+ \Sigma^*)$, where $A\backslash B$ is the left quotient, and $-$ represents the difference of two sets.  In other words, $A\rp B$ is the smallest set $C \subseteq \Sigma^*$ such that $A^* B = A^* C$ and such that no word in $A^+$ is a prefix of any word in $C$.

Example: $\{01\}\rp \{0101010,101\} = \{0,101\}$.

Note that if $A,B$ are regular languages, then $A\rp B$ is also a regular language and can be computed effectively from $A,B$.  (This follows from the standard closure properties of regular languages.)

Definition. The semantics of regular expressions using the $+$ and $@$ operators are defined inductively:


$L(w) = \{w\}$ if $w \in \Sigma^*$
$L(A_1+A_2+\dots+A_k) = L(A_1) \cup L(A_2) \cup \dots \cup L(A_k)$
$L(A^@ B) = \{ab : a \in L(A^*), b \in L(B), b \notin L(A^+ \Sigma^*)\}$
$L(w B) = \{wb : b \in L(B)\}$ if $w \in \Sigma^*$
$L(A^@) = L(A^*)$
$L((A_1+A_2)B) = L(A_1B + A_2 B)$, $L(A(B_1+B_2)) = L(AB_1 + AB_2)$
$L(A_1 A_2 \cdots A_k) = L(A_1 (A_2 \cdots A_k))$


(As a consequence, every regular expression using $+,@$ can be equivalently written as a sum of terms of the form $A_1 A_2 \cdots A_k$ where each $A_i$ is either of the form $w_i$ or $C_i^@$ for some $w_i \in \Sigma^*$ and some regular expression $C_i$ using $+,@$.  Once it is rewritten in that sum-of-products form, the semantics of the resulting expression is given by the definition above.)

For instance, we have the following property: if there is no word in $L(A^+)$ that is a prefix of any word in $L(B)$, then $L(A^@ B) = L(A^* B)$.

In what follows, I won't try to distinguish between a regular expression $E$ and its corresponding language $L(E)$.

Now we can prove a generalization of Arden's lemma.

Lemma 1.  Let $A,B \subseteq \Sigma^*$ be given, and assume $\epsilon \notin A$ and no word in $A^+$ is a prefix of any word in $B$.  Suppose we have the equation $X=AX + B$ (i.e., $X=AX \cup B$).  Then the least solution to this equation is $X=A^@ B$.

Proof.  Arden's lemma says the least solution is $X=A^* B$.  Now, since no word in $A^+$ is a prefix of any word in $B$, in fact $A^* B = A^@ B$, as there can be no ambiguity about how much the $*$ operator gobbles up.

Lemma 2. Let $A,B \subseteq \Sigma^*$ be given, and assume $\epsilon \notin A$.  Suppose we have the equation $X=AX + B$ (i.e., $X=AX \cup B$).  Then the least solution to this equation is $X=A^@ C$ where $C=(A\rp B)$.

Proof.  Arden's lemma says the least solution is $X=A^* B$.  As noted above, $A^* B = A^* C$.  Moreover, no word in $A^+$ is a prefix of any word in $C$, so $A^* C = A^@ C$.

Now given any regular language $L$, we can apply Brzozoswki's method to it, but using Lemma 2 above instead of Arden's lemma.  It's tempting to hope that the result will be a regular expression for $L$ that uses only the operators $+$ and $@$, but unfortunately there's a gap: I don't know whether one can prove that $A\rp B$ can be represented as a regular expression using only the $+$ and $@$ operators.  So, this proof method has a big gaping hole in it.

But perhaps someone will see a way to build on this and prove the desired result.