Matrix Differential equation x'(t) = Ax(t)+b solution defined for non-invertible values

The general solution is the general homogeneous solution plus a particular solution. A very convenient particular solution is a stationary solution. In this problem, a stationary solution $x^*$ is any solution to $Ax^*=-b$. If a stationary solution exists, then the general solution is $x=e^{At}y+x^*$ as $y$ ranges over all of $\mathbb{R}^n$. Here in general $y \neq x(0)$. However, $e^{At}$ is always invertible. Therefore you can always find $y$ to fit to whatever initial condition you want.

This approach can be viewed as being equivalent to what we usually do in the nonnlinear case: $x'=Ax+b$ is equivalent to $(x-x^*)'=A(x-x^*)$, so you solve for $x-x^*$ and add $x^*$ back at the end.

Whether a stationary solution exists or not, and in fact whether the forcing is constant or not, a solution to $x'(t)=Ax(t)+f(t)$ is

$$x(t)=\int_0^t e^{As} f(t-s) ds.$$

In this case $f(t-s) \equiv b$, so we have

$$x(t)=\int_0^t e^{As} b ds.$$

Thus overall the general solution to $x'=Ax+b$ is

$$x(t)=e^{At}y+\int_0^t e^{As}b ds$$

as $y$ ranges over all of $\mathbb{R}^n$. Note that in this form actually $y=x(0)$, which is not the case for the form from the first paragraph.