Converting DFA to RE with Arden's Rule
Your manipulation where you went from $Q_1 = (b \cup ab)Q_2 \cup aaQ_1$ to $Q_1 = (aa)^*(b \cup ab)Q_2$ was not valid.  Instead, in this case I suggest that you first start by finding a nice form for $Q_2$, before trying to manipulate $Q_1$.

A good way to handle $Q_2$ is to note that $Q_2 = a Q_2 \cup b Q_2 \cup \epsilon$ implies $Q_2 = (a \cup b) Q_2 \cup \epsilon$ (by the distribute law); when then apply Arden's rule to conclude that $Q_2 = (a \cup b)^* \epsilon = (a \cup b)^*$.

Now plug in this expression for $Q_2$ into your equations for $Q_0$ and $Q_2$.  You obtain

$$\begin{align*}
Q_0 &amp;= b (a \cup b)^* \cup a Q_1\\
Q_1 &amp;= b (a \cup b)^* \cup a Q_0
\end{align*}$$

Next, take the expression for $Q_1$ and plug it into the first equation.  We see that

$$Q_0 = b (a \cup b)^* \cup a b (a \cup b)^* \cup a a Q_0.$$

Now applying Arden's rule tells us that

$$Q_0 = (aa)^* (b \cup ab) (a \cup b)^*.$$

That's the regular expression that you ultimately wanted.