No, your $L$ (which is sometimes called $\text{REC}$) is $\Sigma_3^0$-complete. However, this proof is somewhat technical. If you just want to show that it and its complement is not c.e., there is a fairly easy proof:
It well known that $K = \{e : \Phi_e(e) \downarrow\}$ is c.e. and but not computable. ($K$ is a variant of the Halting Problem.) It is easy to show that if a set and its complement is c.e., then it must be computable. Hence the complement of $K$, denoted $\bar{K}$, is not c.e.
Define
$\Psi(e,x) = \begin{cases}
0 & \quad \Phi_{e,x}(e)\uparrow \\
0 & \quad \Phi_{e,x}(e) \downarrow \wedge (\exists s)\Phi_{x,s}(x)\downarrow \\
\uparrow & \quad \Phi_{e,x}(e) \downarrow \wedge \Phi_{x}(x)\uparrow
\end{cases}$
$\Psi$ is clearly a partial computable function. By the s-m-n theorem, there exists a computable $f$ such that $\Psi(e,x) = \Phi_{f(e)}(x)$. Thus if $e \in \bar{K}$, then $\Psi(e,x) = \Phi_{f(e)}(x) = 0$ for all $x$ so $W_{f(e)} = \omega$. $\omega$ is computable. So $f(e) \in L$. If $x \in K$, then there is a least $M$ such that $\Phi_{e,M}(e) \downarrow$. So then for all $x > M$, $\Psi(e,x)\downarrow = \Phi_{f(e)}(x)\downarrow$ if and only if $x \in K$. So $W_{f(e)}$ is equal to $K$ except for a finite difference. Since $K$ is not computable, $W_{f(e)}$ is also not computable. Hence $f(e) \notin L$. Thus it has been shown $f$ is the reduction of $\bar{K} \leq_m L$. Since $\bar{K}$ is not c.e., $L$ can not be c.e.
To show that the complement $\bar{L}$ is not c.e. Use
$$\Psi(e,x) = \begin{cases}
0 & \quad \Phi_{e,x}(e) \uparrow \wedge (\exists s)\Phi_{x,s}(x)\downarrow \\
\uparrow & \quad \Phi_{e,x}(e)\uparrow \wedge \Phi_x(x)\uparrow \\
0 & \quad \Phi_{e,x}(e)\downarrow
\end{cases}$$
By a similar argument as above, $\bar{K} \leq_m \bar{L}$. Hence $\bar{L}$ can not be c.e. either.
It has been shown that $L$ and $\bar{L}$ is not c.e.
The basic idea is: The set $K = \{\langle M \rangle : M(\langle M \rangle) \text{ halt }\}$ is recognizable but not decidable. Hence its complement $\bar{K} = \{\langle M \rangle : M(\langle M \rangle) \text{ does not halt}\}$ is not recognizable.
Let $M_x$ denote the Turing machine with code $x$. You want to construct a Turing machine $U$ that reduces $\bar{K}$ to $L$. Given $M$ define $T_M$ to the be Turing machine such that:
$T_M(x)$ halts if $M(\langle M \rangle)$ does not halt by stage $x$
$T_m(x)$ halts if $M(\langle M \rangle)$ halts by stage $x$ and $M_x(x)$ eventually halts
$T_M$ does not halt otherwise
Let $U(\langle M \rangle) = \langle T_M \rangle$
Then if $\langle M \rangle \in \bar{K}$, $T_M$ halts on all input. Hence $L(T_M)$ is decidable. So $U(M) = \langle T_M \rangle \in L$ If $\langle M \rangle \notin \bar{K}$, then $L(T_M)$ agrees with $K$ except on a finite set. Since $K$ is not decidable, $L(T_M)$ is not decidable. So $U(M) = \langle T_M \rangle \notin L$. This give the reduction of $\bar{K} \leq L$. Since $\bar{K}$ is not recognizable, $L$ is not either.
Using a similar idea, you can reduce $\bar{K} \leq \bar{L}$, where $\bar{L}$ is the complement of $L$.
Whenever you approach one of these problems, the first thing to do is what Rogers calls a Tarski-Kuratowski computation: find the level of the arithmetical hierarchy where the set in question lives. In this case, we have that $e \in \text{Tot}$ if for every $n$ there exists an $s$ with $\phi_{e,s}(n)\downarrow$ (that is, $\phi_e(n)$ halts in $s$ steps). Therefore Tot is $\Pi^0_2$ and its complement is $\Sigma^0_2$.
So how can we diagonalize against Tot to show it is not computably enumerable? Well, we can use the leading $\forall$ quantifier in its definition, and the fact that the halting set $K$ is already $\Sigma^0_1$, to show that if Tot was $\Sigma^0_1$ then the halting set would be computable. Recall that, in general, a set is computable if both it and its complement are $\Sigma^0_1$, which is equivalent to the set and its complement both being $\Pi^0_1$.
Let $e$ be given. We want to tell whether $e \in K$, that is, whether $\phi_e(0)$ halts. Let $g$ be such that $g(s) = 0$ if $\phi_{e,s}(0)\uparrow$ and $g(s)$ diverges if $\phi_{e,s}(0)\downarrow$. Notice that $g$ is a partial computable function, and $g$ is total if and only if $\phi_e(0)\uparrow$, and we can compute an index $\gamma_e$ for $g$ uniformly from an index $e$. Therefore, we have
$$
e \in K \leftrightarrow (\exists s)[\phi_{e,s}(0) \downarrow]\\
e \not \in K \leftrightarrow \gamma_e \in \text{Tot}
$$
So, if Tot was computably enumerable, the two facts there the would show that $K$ is computable, which is impossible. The method here was to leverage the $\forall$ quantifier in Tot; if we had a way to decide membership in Tot, that gives us a "free pass" to decide that universal quantifier. The definition of $\gamma_e$ leverages the universal quantifier to tell whether $\phi_e(0)$ does not halt.
How can we show that Tot is not $\Pi^0_1$? Well, we can use the inner $\exists$ quantifier in the definition of Tot, and the fact that the complement $\bar K$ of the halting problem is not computable. In the first part we used Tot to tell whether $\phi_e(0)\uparrow$. Now we will use Tot to tell whether $\phi_e(0)\downarrow$. Given $e$, let $h(x) = \mu s [\phi_{e,s}(0) \downarrow]$. Then $h$ is total if and only if $\phi_e(0)\downarrow$, which is equivalent to $e \not \in \bar K$. Also, we can compute an index $\eta(e)$ for $h$ uniformly from $e$. Thus:
$$
e \in \bar K \leftrightarrow (\forall s)[\phi_{e,s}(0)\uparrow]\\
e \not \in \bar K \leftrightarrow \eta(e) \in \text{Tot}
$$
Therefore, if $\text{Tot}$ was $\Pi^0_1$ then $\bar K$ would be computable, which is impossible.
Best Answer
Every unrecognizable language is undecidable, but there are undecidable languages which are recognizable. In fact, the decidable languages are exactly those which are recognizable and co-recognizable (that is, have recognizable complement).
The Halting Problem (the set of codes for machines which halt on their own code as input - $\{e: \Phi_e(e)\downarrow\}$) is an example of a recognizable, but not decidable, language.