$\def\si{\sigma}$
$\def\th{\theta}$
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
As someone who does a lot of computations with SDEs, I can tell you my own personal reason that I use the differential notation: convenience.
For one thing, when using differential notation, you don't have to worry about mixing up dummy variables. $$ df(X_t) = f'(X_t)dX_t + \frac12f''(X_t)d\langle X\rangle_t $$ is simple and easy to read, but if I wrote it in integral form $$ f(X_t)-f(X_0) = \int_0^tf'(X_s)dX_s + \frac12 \int_0^tf''(X_s)d\langle X\rangle_s $$ it both takes much longer to write and requires me to use and keep track of the dummy variable $s$. This makes things much easier when working with multidimensional processes especially.
Another major factor is a very handy dandy abuse of notation: $dX_tdY_t := d\langle X,Y \rangle_t$. Using this abuse of notation, I can "multiply" stochastic differentials using ordinary algebraic manipulation, for example $$ d\langle f(X) \rangle_t=d f(X_t) df(X_t) = (f'(X_t))^2d\langle X\rangle_t + \frac12f'(X_t)f''(X_t)d\langle X_t, \langle X_t\rangle\rangle_t + \frac14(f''(X_t))^2d\langle\langle X \rangle_t \rangle_t $$ $$ = (f'(X_t))^2d\langle X \rangle_t + 0 + 0 $$ which actually gives the correct answer. One should be VERY careful abusing notation like this, but if you know what hypotheses allow you to use the shortcut, it saves a great deal of time in doing computations.
I should note that I am NEVER going to assign meaning to an expression like $dW_t$ on its own. I will ONLY use it in a context where a corresponding integral expression makes sense.