(In a comment you asked for the "general theory")
The ODE $u''(x)+u'(x)=f(x)\qquad ( a where $f$ is continuous on $\ ]a,b[\ $, $\ -\infty\leq a, is an inhomogeneous linear equation of second order, with constant coefficients. The general theory about such equations says the following:
All solutions are defined on all of $\ ]a,b[\ $.
The associated homogeneous ODE $u''+u'=0$ has the solutions $x\mapsto A+Be^{-x}$ with arbitrary constants $A$, $B$. One arrives at these solutions by studying the characteristic polynomial $\chi(\lambda):=\lambda^2+\lambda$ of the given ODE.
The general solution of the given ODE $(*)$ is of the form $u(x)=u_p(x) + A +Be^{-x}\ ,$ where the so-called particular solution $u_p(\cdot)$ can be any single solution of $(*)$ found by whatever means (guesswork, "Ansatz", "variation of constants", Laplace transform, etc.).
When $f$ is a function of the form $x\mapsto x^n e^{\gamma x}$, or a linear combination of such functions, then a $u_p$ can be found by means of linear algebra.
In your example you can substitute $u'(x):=y(x)e^{-x}$ and will then get $y'(x)=f(x)e^x$, or $y(x)=y_0+\int_0^x f(t)e^t\ dt$. This leads to $u'(x)=y_0e^{-x}+\int_0^x f(t)e^{t-x}\ dt$. Integrating once more we have $u(x)=y_1-y_0e^{-x}+\int_0^x\ \int_0^{x'}f(t)e^{t-x'}\ dt\ dx'=y_1-y_0e^{-x}+\int_0^x f(t)\bigl(1-e^{t-x}\bigr)\ dt\ .$ Note that the terms $y_1-y_0e^{-x}$ correspond to $ A +Be^{-x}$ in the above general setup.
But we are not done yet: The given problem may still have $0$, $1$, or an infinity of solutions. This has to do with the somewhat nonstandard boundary conditions. I don't think there is a "general theory" covering exactly your data. Therefore we have to proceed ad hoc.
One of your conditions is $u(0)=u'(0)$. Now $u(0)=y_1-y_0$ and $u'(0)=y_0$. Therefore necessarily $y_1=2y_0$. To satisfy the boundary condition at $L$ one has to collect the $f$-integrals for $u$ and $u'$, and then check whether $y_0$ can be chosen suitably.
If my calculations are correct, the answer is the following: When $\int_0^Lf(t)\ dt=0$ then the problem has infinitely many solutions, otherwise none.