A general SDE look like the following:
$$ \mathrm{d}\psi=a\mathop{}\!\mathrm{d}t+b\mathop{}\!\mathrm{d}W,\tag{1} $$
where $\psi:t\mapsto y = \psi(t)$ is the solution, while $a$ and $b$ can be both, constants or any function, right?
Equation $1$ can be rewritten as:
$$ \frac{\mathrm{d}\psi}{\mathrm{d}t}=a\mathop{}\!\mathrm{d}t+b\mathop{}\!\mathrm{d}W, $$
or as:
$$ \psi(t)-\psi(t_0)=\int a\mathop{}\!\mathrm{d}t+\int b\mathop{}\!\mathrm{d}W. $$
so it's not clear to me why it's common to use $\mathrm{d}\psi$ instead of $\psi'(t)$ to abbreviate. Also, is $\psi(t_0)$ the integration constant of $\int\frac{\mathrm{d}\psi}{\mathrm{d}t}\mathop{}\!\mathrm{d}t$? If so, why it has the minus sing?
The differential of the function $y = \psi(t)$ is:
$$ \mathop{}\!\mathrm{d}\psi=\psi'(t)\mathop{}\!\mathrm{d}t=\frac{\mathrm{d}y}{\mathrm{d}t}\mathop{}\!\mathrm{d}t, $$
and since the two $\mathop{}\!\mathrm{d}t$ simplify, the final result would be $\mathrm{d}y$? Then in this case, Ito's lemma would help me find the expression of this $\mathrm{d}y$?
In the case of a system of two SDEs, can I just use Ito's lemma on both the equations and then match the initial conditions with the constants?
Indeed, as mentioned in the comments the function
$$\psi(t)=\int_{0}^{t}b_{s}ds+\int_{0}^{t}a_{s}dW_{s}$$
is not differentiable for nonzero $a$ eg. take $a=1$
$$\psi(t)=\int_{0}^{t}b_{s}ds+B_{t}.$$