The sum of $Z$ many draws from the Exponential distribution are distributed Erlang. We can prove this using convolutions by induction, basically iterating on
$$(g \cdot g)(t) = \int_0^t g(s)g(t-s)\, ds $$
My random variable is related to Exponential: its CDF is given by
$$ F(x) = e^{\lambda(x - x_1)}$$
and it is defined over $[x_0, x_1]$ with $x_0 \geq 0$. It will always be the case that $F(x_0) > 0$, implying that there is a mass point at $x_0$. In other words, $F$ is non-differentiable at $x_0$.
I would like to compute the distribution that the sum of $j$ many draws from $F$ follows. How can I proceed here?
Not sure if this will lead anywhere, but here is one idea.
Rather than dealing with a sum of $j$ draws, let's deal with the sum of $2$ draws and see if we can generalize the result. So let $S = X_1+X_2$, where $X$ has the cdf $F_X(x)$ which you have described above. Then, the distribution of $S$ is characterized by its cdf, $$ F_S(s) = \mathbb{P}[S \le s] = \mathbb{P}[X_1+X_2 \le s]. $$ The last expression can be evaluated by conditioning on $X_2$. Since you have point weights, you can use Riemann-Stjeltjes integration to write $$ F_S(s) = \mathbb{P}[X_1+X_2 \le s] = \int_{x_0}^{x_1} \mathbb{P}[X_1+X_2 \le s|X_2 = t]dF_X(t) $$ which you can now simplify and express the inner probability as a function of $F_X(\cdot)$ and hope the resulting integral is reasonably easily computable...