Every state and every item from the set $\{a,b\}$ has a transition to another state, and every state is an accepting state. Therefore, all states are equivalent and the regular expression (in your notation) is just $(a ∪ b)^*$.
The automaton in question is equivalent to the following Kleene Algebraic fixed-point system of inequations:
$$A ≥ a A + b B + 1,\quad B ≥ a B + b B + 1,\quad A,$$
with the last item, $A$, indicating the starting expression. The least fixed point solution for $(A,B)$, substituted into the starting expression $A$ yields the regular expression in question.
As you noted, you can factor $B$:
$$A ≥ a A + b B + 1,\quad B ≥ (a + b) B + 1,\quad A.$$
You can solve for $B$. In general, the least fixed point solution to $x ≥ u x + v$, in $x$, which is denoted $μx·(ux + v)$, in Kleene algebra, is just $x = u^* v$. Thus, the reduction:
$$A ≥ a A + b B + 1,\quad B ≥ (a + b)^*,\quad A.$$
Upon substitution for $B$ and elimination, this reduces to:
$$A ≥ a A + b (a + b)^* + 1,\quad A.$$
In a similar way, we find the solution for $A$:
$$A ≥ a^* (b (a + b)^* + 1),\quad A,$$
and, upon substitution and elimination of $A$:
$$a^* (b (a + b)^* + 1).$$
(Your mistake, as already noted in another reply, is that you had the equivalent of $a^* b (a + b)^* + 1$, instead of $a^* (b (a + b)^* + 1)$.)
Kleene-algebraic identities can be used, particularly $(u + v)^* = u^* (v u^*)^*$ and $u^* = u u^* + 1$, reducing this to
$$\begin{align}
a^* (b (a + b)^* + 1)
&= a^* (b a^* (b a^*)^* + 1)\\
&= a^* (b a^*)^*\\
&= (a + b)^*,
\end{align}$$
or, in your notation $(a ∪ b)^*$.
The axioms cited in the linked article are sufficient to prove any and all Kleene-algebraic identities. They are equivalent to the following set of properties:
$$
(x + y) + z = x + (y + z),\quad x + 0 = x = 0 + x,\quad x + y = y + x,\quad x + x = x,\\
(xy)z = x(yz),\quad x 1 = x = 1 x,\quad p 0 q = 0,\quad p (x + y) q = p x q + p y q,\\
0^* = 1,\quad μx·(a + bx + xc) = b^* a c^*.
$$
(Edit: I put in $0^* = 1$, too, otherwise you have to split the least-fixed-point equation into $μx·(a + bx) = b^* a$ and $μx·(a + xc) = a c^*$.)
These axiomatizations come from 1994. Many texts - even now - are obsolete, still using frameworks from the 1960's and not including these more recent developments, which themselves are now over a generation past.
If you strengthen it by replacing the last axiom with
$$p (μx·f(x)) q = \sup_{n≥0} p f^n(0) q,$$
where
$$f^0(x) = x,\quad f^1(x) = f(x),\quad f^2(x) = f(f(x)),\quad f^3(x) = f(f(f(x))),\quad ⋯,$$
where $f(x)$ denotes an arbitrary Kleene-polynomial in $x$, then you get an axiomatization for fixed-point-closed Kleene algebras that is complete with respect to the algebra of context-free subsets; i.e. an algebra for "context-free" expressions: μ-continuous Chomsky Algebras.
This comes from the mid-2010's (2014, I believe) and was in 2018 proven equivalent to another axiomatization that arose in 2008 (the algebra of C-dioids), which used in place of the last property that (a) every context-free subset $U$ has a least upper bound $\sup U$, and (b) $\sup \{p\}U\{q\} = p (\sup U) q$, for context-free subsets $U$.
Can you detail how you solved the system?
– Shaull Mar 01 '13 at 17:35