In the book "An introduction to PDES", by Pinchover and Rubinstein, an example problem is
$$u_x =c_0 u(x,y) + c_1(x,y)$$
where $c_0$ is a constant and the initial value
$$u(0,y)=y$$
is given. The authors then jump the the "obvious" solution
$$ u(x,y) = e^{c_0 x}\left[\int_0^x e^{-c_0 \xi}c_1(\xi,y)d\xi + y\right]$$
Is it true that the solution without the initial value is
$$ u(x,y) = e^{c_0 x} \int e^{-c_0x}c_1(x,y) dx + e^{c_0x}f(y)$$ and that the initial value condition can be written $$u(0,y) = \int_{-\infty}^0 e^{-c_0 x} c_1(x,y) dx + f(y) =y$$ and hence we know that $$f(y)=y, \hspace{1cm} \int_{-\infty}^0 e^{-c_0x}c_1(x,y) dx = 0$$ and that is why we can include those limits in the solution-integral ?
It is true, but you have some unnecessary freedom in your solution. You have an arbitrary constant of integration for the first term (which is a particular solution of the equation) plus an arbitrary function in the second term, which represents the general solution to the homogeneous equation. It would be equally general to say that the general solution is $$ u(x,y)=e^{c_0x}f(y)+e^{c_0x}\int\limits_0^xe^{-c_0s}c_1(s,y)\,ds $$ and then your initial condition implies $f(y)=y$. Of course you can replace the lower limit in the integral by any real number (that changes your $f(y)$ by some constant). Setting it equal to $-\infty$ is not safe because you do not know whether the improper integral converges (this will depend on the behavior of $c_1(\cdot, y)$) In any case, you cannot set the integral $$ \int\limits_{-\infty}^0e^{-c_0s}c_1(s,y)\,ds $$ equal to zero, since all the functions are given and fixed. All you have to play with is $f(y)$, which is equal to $y$ if you set the lower limit in the integral equal to zero.