[proofplan]
We first show that $A'$ is computably enumerable relative to $A$ by using the oracle $A$ to dovetail all diagonal computations $\varphi_e^A(e)$. For completeness, let $B$ be any set computably enumerable in $A$, choose an oracle index $e$ enumerating it as a domain, and uniformly transform each input $x$ into an index for a program that ignores its own input and simulates $\varphi_e^A(x)$. This computable index transformation sends $x$ into $A'$ exactly when $x$ belongs to $B$, giving the required many-one reduction.
[/proofplan]
[step:Enumerate the diagonal halting set relative to $A$]
Let $U^A: \mathbb{N} \times \mathbb{N} \rightharpoonup \mathbb{N}$ denote the universal partial $A$-computable function associated to the fixed acceptable enumeration, so that
\begin{align*}
U^A(e,x) = \varphi_e^A(x)
\end{align*}
whenever either side is defined.
Define an oracle enumeration procedure $E^A: \mathbb{N} \rightharpoonup \mathbb{N}$ as follows. At stage $s \in \mathbb{N}$, simulate the first $s$ steps of each computation
\begin{align*}
\varphi_0^A(0),\ \varphi_1^A(1),\ \dots,\ \varphi_s^A(s),
\end{align*}
using the oracle $A$ to answer all oracle queries made during these finite simulations. Whenever the simulation of $\varphi_i^A(i)$ is observed to halt for the first time by stage $s$, enumerate $i$.
This procedure is computable relative to $A$ because each finite stage contains only finitely many finite oracle computations, and every oracle query is answered by membership in $A$. If $i \in A'$, then $\varphi_i^A(i)$ halts after some finite number of computation steps, so some stage eventually enumerates $i$. If the procedure enumerates $i$, then a finite simulation has witnessed that $\varphi_i^A(i)$ halts, hence $i \in A'$. Therefore the range enumerated by $E^A$ is exactly $A'$, so $A'$ is computably enumerable in $A$.
[guided]
We must prove that $A'$ is computably enumerable using oracle $A$. The set $A'$ consists of exactly those indices whose own diagonal oracle computation eventually halts:
\begin{align*}
A' = \{e \in \mathbb{N} : \varphi_e^A(e) \downarrow\}.
\end{align*}
Let $U^A: \mathbb{N} \times \mathbb{N} \rightharpoonup \mathbb{N}$ be the universal partial $A$-computable function for the fixed acceptable enumeration. This means that, for every $e,x \in \mathbb{N}$,
\begin{align*}
U^A(e,x) = \varphi_e^A(x)
\end{align*}
whenever the computation is defined.
The enumeration procedure is the standard dovetailing procedure, but every simulated computation is allowed to ask oracle questions about $A$. At stage $s \in \mathbb{N}$, run the first $s$ steps of each of the finitely many computations
\begin{align*}
\varphi_0^A(0),\ \varphi_1^A(1),\ \dots,\ \varphi_s^A(s).
\end{align*}
If the computation $\varphi_i^A(i)$ is seen to halt at this stage, and $i$ has not already been listed, enumerate $i$.
Why is this an $A$-computable enumeration? At each stage there are only finitely many machine steps to simulate. Whenever one of those finite simulations asks whether a natural number $n$ belongs to $A$, the oracle supplies the answer. Hence the entire stage is effective relative to $A$.
Now verify that the enumeration has exactly the desired range. If $i$ is listed, then the procedure has actually observed a finite halting computation of $\varphi_i^A(i)$, so $\varphi_i^A(i) \downarrow$ and therefore $i \in A'$. Conversely, if $i \in A'$, then $\varphi_i^A(i)$ halts after some finite number $t \in \mathbb{N}$ of computation steps. At every stage $s \ge \max\{i,t\}$, the dovetailing procedure has simulated enough of $\varphi_i^A(i)$ to detect this halt, so $i$ is eventually enumerated. Thus the enumerated set is exactly $A'$.
[/guided]
[/step]
[step:Convert an arbitrary relative enumeration into a diagonal halting problem]
Let $B \subseteq \mathbb{N}$ be computably enumerable in $A$. By definition, there exists an index $e \in \mathbb{N}$ such that
\begin{align*}
B = \{x \in \mathbb{N} : \varphi_e^A(x) \downarrow\}.
\end{align*}
For each $x \in \mathbb{N}$, define an oracle program $P_x^A: \mathbb{N} \rightharpoonup \mathbb{N}$ by
\begin{align*}
P_x^A(y) =
\begin{cases}
0, & \text{if the simulation of } \varphi_e^A(x) \text{ halts},\\
\uparrow, & \text{if the simulation of } \varphi_e^A(x) \text{ does not halt},
\end{cases}
\end{align*}
where $y \in \mathbb{N}$ is ignored. The construction of $P_x^A$ from $x$ is effective without using the oracle: it writes a program with the fixed constants $e$ and $x$, ignores its input, and then runs the universal oracle simulation of $\varphi_e^A(x)$. Therefore there is a total computable function
\begin{align*}
f: \mathbb{N} &\to \mathbb{N}
\end{align*}
such that, for every $x,y \in \mathbb{N}$,
\begin{align*}
\varphi_{f(x)}^A(y) = P_x^A(y).
\end{align*}
In particular, substituting $y = f(x)$ gives
\begin{align*}
\varphi_{f(x)}^A(f(x)) \downarrow
\iff
P_x^A(f(x)) \downarrow
\iff
\varphi_e^A(x) \downarrow.
\end{align*}
Using the definition of $B$ and the definition of $A'$, this becomes
\begin{align*}
x \in B \iff f(x) \in A'.
\end{align*}
Since $f$ is total computable, this proves $B \le_m A'$.
[/step]
[step:Conclude many-one completeness]
The first step proves that $A'$ is computably enumerable in $A$. The second step proves that every set $B \subseteq \mathbb{N}$ computably enumerable in $A$ admits a total computable reduction $f: \mathbb{N} \to \mathbb{N}$ satisfying
\begin{align*}
x \in B \iff f(x) \in A'
\end{align*}
for every $x \in \mathbb{N}$. Hence $A'$ is many-one complete among the sets computably enumerable in $A$.
[/step]
