$\def\si{\sigma}$
$\def\th{\theta}$
Let $M(u,t)=\int_0^t\si(u,r)\,dW_r$ and assume the sample paths of $\si$ have continuous partial derivatives. In this answer, I will repeatedly use the fact that if $\th$ is a bounded variation process, then
$$
\int_0^t\th_r\,dW_r=\th_tW_t-\int_0^tW_r\,d\th_r.\tag{IBP}
$$
Lemma.
$$
\int_0^s M(u,t)\,du = \int_0^s\int_0^t\si(u,r)\,dW_r\,du
= \int_0^t\int_0^s\si(u,r)\,du\,dW_r.
$$
Proof. Fix $s$. Let $Y_t=\int_0^t\int_0^s\si(u,r)\,du\,dW_r$. By (IBP), we have
$$
Y_t = W_t\int_0^s\si(u,t)\,du - \int_0^tW_r\int_0^s\si_2(u,r)\,du\,dr,
$$
where $\si_2$ is the partial derivative of $\si$ with respect to the second argument. For each sample path, the above integrals are ordinary integrals. Thus, by Fubini's theorem,
\begin{align}
Y_t &= W_t\int_0^s\si(u,t)\,du - \int_0^s\int_0^tW_r\si_2(u,r)\,dr\,du\tag{1}\\
&= \int_0^s\left({W_t\si(u,t) - \int_0^tW_r\si_2(u,r)\,dr}\right)\,du.
\end{align}
By (IBP),
$$
\int_0^t W_r\si_2(u,r)\,dr = W_t\si(u,t) - \int_0^t \si(u,r)\,dW_r.\tag{2}
$$
Thus, $Y_t=\int_0^s\int_0^t \si(u,r)\,dW_r\,du$. $\square$
Let $X_t=\int_0^t M(u,t)\,du$. By (1) with $s=t$,
$$
X_t = W_t\int_0^t\si(u,t)\,du - \int_0^t\int_0^tW_r\si_2(u,r)\,dr\,du.
$$
Thus, again using (IBP),
\begin{align}
dX_t &= \int_0^t\si(u,t)\,du\,dW_t
+ W_t\left({
\si(t,t) + \int_0^t\si_2(u,t)\,du
}\right)\,dt\\
&\qquad - \left({
\int_0^tW_r\si_2(t,r)\,dr + \int_0^tW_t\si_2(u,t)\,du
}\right)\,dt\\
&= \int_0^t\si(u,t)\,du\,dW_t
+ \left({
W_t\si(t,t) - \int_0^tW_r\si_2(t,r)\,dr
}\right)\,dt.
\end{align}
By (2) with $u=t$, this gives
\begin{align}
dX_t &= \int_0^t\si(u,t)\,du\,dW_t
+ \int_0^t \si(t,r)\,dW_r\,dt\\
&= \int_0^t\si(u,t)\,du\,dW_t
+ M(t,t)\,dt.
\end{align}
Finally, let
\begin{align}
L(u,t) &= L(u,0) + \int_0^t\mu(u,r)\,dr + \int_0^t\si(u,r)\,dW_r\\
&= L(u,0) + \int_0^t\mu(u,r)\,dr + M(u,t).
\end{align}
Then
\begin{align}
\int_0^t L(u,t)\,du &= \int_0^t L(u,0)\,du
+ \int_0^t\int_0^t\mu(u,r)\,dr\,du + \int_0^t M(u,t)\,du\\
&= \int_0^t L(u,0)\,du + \int_0^t\int_0^t\mu(u,r)\,dr\,du + X_t.
\end{align}
Therefore,
\begin{align}
d\left({\int_0^t L(u,t)\,du}\right)
&= L(t,0)\,dt + \int_0^t\mu(t,r)\,dr\,dt + \int_0^t\mu(u,t)\,du\,dt\\
&\qquad + \int_0^t\si(u,t)\,du\,dW_t + M(t,t)\,dt\\
&= L(t,t)\,dt + \int_0^t\mu(u,t)\,du\,dt
+ \int_0^t\si(u,t)\,du\,dW_t.
\end{align}
This formula has an extra $dW$ term which your formula does not. In hindsight, this seems reasonable. By (1), $\int_0^s M(u,t)\,du$ is a process such that if we fix $t$, then it is bounded variation in $s$, but if we fix $s$, then it is quadratic variation in $t$. Your formula implies that this process is bounded variation on the diagonal $s=t$, which seems fairly counterintuitive. (Edit: I just noticed the phrase, "not including any terms involving the brownian motion", so I guess the formula you posted at the end was intentionally incomplete.)
Edit 2: If we interpret expressions such as $\int_0^t f(u,t)\,dZ_t\,du$ as meaning $\left({\int_0^t f(u,t)\,du}\right)\,dZ_t$, then we have just proved that under suitable assumptions on $\mu$ and $\si$,
$$
d\left({\int_0^t L(u,t)\,du}\right)
= L(t,t)\,dt + \int_0^t dL(u,t)\,du.
$$
Best Answer
We will apply Ito's formula to the process $e^{at}X_t$. Let $f(t,x) := e^{at}x$, so by Ito's formula \begin{align*} d(e^{at}X(t)) = df(t,X(t)) &= \partial_t f(t,X_t)dt + \partial_x f(t,X_t)dX(t) + \frac 12 \partial_{xx}f(t,X_t)dX(t)dX(t) \\ &= ae^{at}X(t)dt + e^{at}(-aX(t)dt + \sigma dW(t)) \\ &= \sigma e^{at}dW(t). \end{align*} Re-writing in integral form, this says $$e^{at}X(t) = \int_0^t \sigma e^{as}dW(s) $$ so solving for $X(t)$ gives $$X(t) = e^{-at}\int_0^t\sigma e^{as}dW(s) = \sigma \int_0^t e^{-a(t-s)}dW(s).$$