Replacing your $\{0,1\}$, respectively by $\{a,b\}$ and using more algebraic notation with $1$ in place of $ε$, the automaton can be presented either as a right-affine system
$$
A ≥ a A + b B + 1,\quad
B ≥ a C + b B,\quad
C ≥ a A + b B,\quad
A
$$
or a left-affine system
$$
A ≥ A a + C a + 1,\quad
B ≥ A b + B b + C b,\quad
C ≥ B a,\quad
A
$$
where, in both cases, the last item refers to the "starting expression", following the remainder of the system, which is a fixed-point system of inequations in $(A,B,C)$ whose solution is to be applied to the starting expression. The solution is chosen as the least fixed point solution over a Kleene algebra. That's the contemporary framing of the problem (i.e. inequations, not equations).
Apart from that, everything else you said applies. Your methods - worked out on the left-affine system - yield
$$(a + b (b + a b)^* a a)^*,\quad (a + b b^* a (b b^* a)^* a)^*,$$
and you forgot the Kleene-star on the second expression, as your diagram indicates should be there.
All Kleene algebra identities can be derived from a small set of equational properties (namely: those for an idempotent semiring, a.k.a. "dioid" as it is sometimes also known as) ... plus two conditional properties that can be stated as follows:
$$μx·(a + b x) = b^* a,\quad μx·(a + x c) = a c^*,$$
where $μx·φ(x)$ denotes the least fixed point solution to $x ≥ φ(x)$. You could call those the "Arden Equations", if you wish. They have no official name. All finite one-sided affine systems have least fixed point solutions - under these axioms. And all properties of the algebras of regular expressions (which are known as the "free Kleene algebras") can be derived from these axioms.
For your case, you have the identity $(u + v)^* = u^* (v u^*)^*$. Thus
$$(b + a b)^* = b^* (a b b^*)^*.$$
From this, it follows that
$$a + b (b + a b)^* a a = a + b b^* (a b b^*)^* a a.$$
Then, you have the identity $(u v)^* u = u (v u)^*$, from which it follows that
$$b b^* (a b b^*)^* a = b b^* a (b b^* a)^*.$$
Thus,
$$a + b (b + a b)^* a a = a + b b^* a (b b^* a)^* a.$$
This shows that your two expressions are equal.
Using distributivity $u (v + w) = u v + u w$, you can proceed to write
$$a + b b^* (a b b^*)^* a a = (1 + b b^* a (b b^* a)^*) a.$$
From the property $1 + u u^* = u^*$, it follows that
$$(1 + b b^* a (b b^* a)^*) a = (b b^* a)^* a.$$
Thus, your expression can also be written as
$$((b b^* a)^* a)^*.$$
Using the property $(u^* v)^* = 1 + (u + v)^* v$ (which can be derived from the ones already listed), it follows that
$$((b b^* a)^* a)^* = 1 + (b b^* a + a)^* a.$$
Using distributivity on the other side $(u + v) w = u w + v w$ and the other properties already cited, we have
$$(b b^* a + a)^* = ((b b^* + 1) a)^* = (b^* a)^* = 1 + (a + b)^* a,$$
which also used commutativity $a + b = b + a$.
Thus
$$((b b^* a)^* a)^* = 1 + (b b^* a + a)^* a = 1 + a + (a + b)^* a a.$$
For the right-affine system you can reverse-substitute $C$ in for $a A + b B$ to get:
$$
A ≥ C + 1,\quad
B ≥ a C + b B,\quad
C ≥ a A + b B,\quad
A.
$$
Nobody ever said you always had to eliminate variables. You can put them back in, too.
Now, you can eliminate $A$, applying distributivity (and commutativity)
$$
B ≥ a C + b B,\quad
C ≥ a + a C + b B,\quad
1 + C.
$$
Then you can reverse-substitute $B$ for $a C + b B$ to get
$$
B ≥ a C + b B,\quad
C ≥ a + B,\quad
1 + C.
$$
Then, you can forward-substitute and eliminate $C$ (and apply distributivity) to get
$$
B ≥ a a + (a + b) B,\quad
1 + a + B.
$$
Finally, use the Arden axiom on $B$ to get
$$
B ≥ (a + b)^* a a,\quad
1 + a + B.
$$
and substitute out $B$:
$$1 + a + (a + b)^* a a.$$
For the left-affine system, if you first factor, you get:
$$
A ≥ (A + C) a + 1,\quad
B ≥ (A + B + C) b,\quad
C ≥ B a,\quad
A.
$$
This shows that the actual variables are not $(A,B,C)$, but
$$(A, D, E) = (A, A + C, A + B + C).$$
Reverse-substitute for $A$ and $B$ in $D$, and for $A$, $B$ and $C$ in $E$, and use commutativity to get
$$
A ≥ 1 + D a,\quad
B ≥ E b,\quad
C ≥ B a,\quad
A.
$$
Then, reverse-substitute on the left-hand side and use commutativity and distributivity to get:
$$
A ≥ 1 + D a,\quad
D ≥ 1 + D a + B a = 1 + (B + D) a = 1 + E a,\\
E ≥ 1 + E a + E b = 1 + E (a + b),\quad
A.
$$
Forward-substitute for and eliminate $A$:
$$
D ≥ 1 + E a,\quad
E ≥ 1 + E (a + b),\quad
1 + D a.
$$
Then, forward substitute for and eliminate $D$ and apply distributivity
$$
E ≥ 1 + E (a + b),\quad
1 + a + E a a.
$$
Apply the Arden axiom on $E$
$$
E ≥ (a + b)^*,\quad
1 + a + E a a.
$$
Then forward-substitute for and eliminate $E$:
$$1 + a + (a + b)^* a a.$$
If "simplicity" is measured by star-height, then this is optimial. It has star-height 1, while the other expressions have star-height 2. The problem of minimizing star-height is highly non-trivial, but solved (in the 1990's).
You can go much further with everything ... nowadays.
To get an algebra suited for context-free expressions it suffices to replace the Arden axioms by the μ-continuity/distributivity axiom
$$u (μx·φ(x)) v = \sup_{n≥0} u φ^n(0) v,$$
where
$$φ^0(x) = x,\quad φ^1(x) = φ(x),\quad φ^2(x) = φ(φ(x)),\quad φ^3(x) = φ(φ(φ(x))),\quad ⋯,$$
and $φ(x)$ is an arbitrary Kleene-polynomial in $x$. Solutions can be expressed Kleene-algebraically, using context-free expressions, e.g.
$$μx·(u x v + w) = b (u p + q v)^* d,$$
which - here - extends the Kleene algebra given by $X = \{u,v,w\}$ by the inclusion of the "bracket operators" $Z = \{b,d,p,q\}$ satisfying
$$b d = 1 = p q,\quad b q = 0 = p d,\quad x z = z x\quad (x ∈ X,\quad z ∈ Z).$$
They are known equivalently as μ-continuous Chomsky algebras / C-dioids the two characterizations having arisen, respectively, in 2013 and 2008, their equivalence proven in 2018. Their representation as Kleene-algebras with the bracket operators affixed is via a result, (finally and belatedly) put into the peer review literature a few years ago, that is an algebraic generalization of, and a completion of, the Chomsky-Schützenberger Theorem.
So, similar and more elaborate exercises can be used to crunch non-linear fixed-point systems (e.g. systems that correspond to context-free grammars, or translation grammars such as arise in parsing theory) into context-free expressions.
The worst cases, if you allow the sharing of sub-expressions, are cubic in the number of variables for turning one-sided affine systems into regular expressions; but linear (and star-height 1) - for both the affine and more general non-linear systems, if you use the bracket operators $\{b,d,p,q\}$.
If you allow for commutativity of the product - the situation that occurs with Parikh's Theorem - then the Arden axioms, alone, suffice and you can write $μx·φ(x) = φ'(φ(0))^* φ(0)$, where the usual rules of differentiation on polynomials (after noting, for instance, that $d(x^2)/dx = 2x = x$, since $x + x = x$ in Kleene algebras) is extended to the Kleene star with $d(x^*)/dx = x^*$. For commutative Kleene algebras, the star behaves like the exponential: $u^* v^* = (uv)^*$ and $0^* = 1$.
