Since the OP has recently asked a related question for a different PDE and since this can all be answered in a completely general fashion, here is a general approach.
General setting.
Let us consider the following general setting: Let $X$ be a Banach space (think of a function space), let $A: X \supseteq \operatorname{dom}(A) \to X$ be a generator of a $C_0$-semigroup $(S(t))_{t \ge 0}$ on $X$ (think of $A$ is a "sufficiently good" differential operator), let $T > 0$ be a real number, and let $f \in L^\infty([0,T]; X)$. For every $t \in [0,T]$, let $B(t): X \to X$ be a bounded linear operator on $X$ and assume that $\sup_{t \in [0,T]} \lVert B(t) \rVert < \infty$ and that the mapping $[0,T] \ni t \mapsto B(t)x \in X$ is strongly measurable for every $x \in X$.
We are interested in the initial value problem
$$
\begin{cases}
\dot u(t) = Au(t) + B(t)u(t) + f(t) \quad \text{ for } t \in [0,T], \\
u(0) = u_0
\end{cases}\label{1}\tag{$\ast$}
$$
for $u_0 \in X$, where $u: [0,T] \to X$ is the unknown function.
Mild formulation.
What we will actually study a mild formulation of \eqref{1}, i.e. we will look for a function $u \in L^\infty([0,T];X)$ that satifies the following fixed point problem:
$$\DeclareMathOperator{\Dm}{d\!}
u(t) = S(t)u_0 + \int_0^t S(t-s)\big(B(s)u(s) + f(s)\big) \Dm s
\quad \text{ for } t \in [0,T]. \label{2}\tag{$\ast\ast$}
$$
Here, the integral is meant as a Bochner integral with values in the Banach space $X$.
Well-posedness. Existence and uniqueness of mild solutions follows from the following theorem.
Theorem 1.
Let $u_0 \in X$.
There exists precisely one function $u \in L^\infty([0,T];X)$ that satisfies the equation \eqref{2} and this function $u$ is actually continuous with respect to time.
Moreover, the solution can be obtained by iteration as follows: Consider the mapping $G: L^\infty([0,T];X) \to L^\infty([0,T];X)$ given by
$$
(Gw)(t) = S(t)u_0 + \int_0^t S(t-s)\big(B(s)w(s) + f(s)\big) \Dm s
$$
for $w \in L^\infty([0,T];X)$ and for $t \in [0,T]$.
Then for every $w \in L^\infty([0,T];X)$ the sequence $(G^n w)$ converges (with respect to the sup norm) to $u$ as $n \to \infty$.
Proof.
One first has to check that, for every $w \in L^\infty([0,T];X)$, the integrand of $G$ is strongly measurable, such that the integral actually makes sense.
Next, one can show that $G$ maps the space $L^\infty([0,T];X)$ into the space $C([0,T];X)$ of continuous $X$-valued functions; this is essentially a consequence of the dominated convergence theorem for Bochner integrals. The details are a bit technical, but quite straightforward, so I omit them here. We conclude that, in particular, a fixed point of $G$ is automatically a continuous function with respect to time.
Now we only need to show that one Banach's fixed point theorem can be applied to $G$.
To this end, let $M_S := \sup_{t \in [0,T]} \lVert S(t) \rVert$ and $M_B := \sup_{t \in [0,T]} \lVert B(t) \rVert$.
For $w_1, w_2 \in L^\infty([0,T];X)$ we can show be induction over $n$ that
$$
\lVert (G^n w_1)(t) - (G^n w_2)(t) \rVert_X
\le
\frac{(M_S M_B t)^n}{n!}
$$
for all $t \in [0,T]$ and all integers $n \ge 0$.
Hence, for each such $n$ the mapping $G^n$ is Lipschitz continuous with Lipschitz constant at most $\frac{(M_S M_B \, T)^n}{n!}$.
Since the Lipschitz constants are summable over $n \in \mathbb{N}_0$, Banach's fixed point theorem applies to $G$. $\square$
Remark.
Note that there is no need to restrict anything to small times here. The proof shows that the Picard-Lindelöf iterates $G^n w$ converge globally on $[0,T]$.
Positivity.
For this, the Banach space $X$ needs an order structure. Let $X_+ \subseteq X$ be a closed convex cone and, for $x,y \in X$, set $x \le y$ iff $y-x \in X_+$. Hence, the elements of $X_+$ are precisely those $x \in X$ that satisfy $x \ge 0$.
A bounded linear operator $C: X \to X$ is called positive, which we denote by $C \ge 0$, if $CX_+ \subseteq X_+$. Equivalently, $x \le y$ implies $Tx \le Ty$.
From Theorem 1 one obtains:
Corollary 2.
Assume that the $C_0$_semigroup $(S(t))_{t\ge 0}$ is positive, i.e. that $S(t) \ge 0$ for all $t \ge 0$.
Assume moreover that $f(t) \ge 0$ for all $t \in [0,T]$ and that there exists a real number $c \ge 0$ such that $B(t) + c \operatorname{id}_X \ge 0$ for all $t \in [0,T]$.
If $u_0 \ge 0$ than the solution $u$ of \eqref{2} satisfies $u(t) \ge 0$ for all $t \in [0,T]$.
Proof. First note that the term $Au(t) + B(t)u(t)$ in \eqref{1} can be rewritten as $(A-c \operatorname{id}_X)u(t) + (B(t)+c \operatorname{id}_X)u(t)$. The semigroup generated by $A-c \operatorname{id}_X$ is given by $e^{-ct} S(t)$ at time $t$, so this semigroup is also positive. The advantage is that we have now achieved that the operator $B(t)+c \operatorname{id}_X$ in the non-autonomous term is positive. Hence, we may and shall assume from now on that each of the operators $B(t)$ is positive and that $c=0$.
We order the space $L^\infty([0,T];X)$ by the pointwise almost everyhwere order induced from $X$. Since $u_0 \ge 0$ and as the semigroup $(S(t))_{t \ge 0}$ is positive, the operators $B(t)$ are positive and one also has $f(t) \ge 0$ for all $t$, it follows that $G$ maps the cone of $L^\infty([0,T];X)$ to itself. As this cone is closed, we may start with any function $w$ in this cone and conclude that the limit $u = \lim_{n \to \infty} G^n(w)$ is also in the cone.
Thus, $u(t) \ge 0$ for $t \in [0,T]$. $\square$
Remark.
Once one knows that $u(t) \ge 0$ for all $t \in [0,T]$, one can get a bit more information from the mild equation \eqref{2}.
For instance, if the semigroup $(S(t))_{t \ge 0}$ has somekind of positivity improving property (as is for instance the case for the heat semigroup), then \eqref{2} shows that the same is true for the non-autonomous equation that we are considering.
Applications.
The equation in the question can be treated by this framework: take $A = a \partial_x^2$ and let $B(t)$ be the multiplication with $g(\mathbin{\cdot},t)$.
Note that one does not need the function $g$ to be positive.
Moreover, one does not need strict positivity of the initial value $u_0$; the inequality $u_0 \ge 0$ suffices to get $u(t) \ge 0$ for all $t \in [0,T]$.
Since the framework is very general, one can of course also consider $A$ to be a second order elliptic operator on a multidimensional domain (for instance with Dirichlet or Neumann boundary conditions).
Similarly, the kinetic equation in OP's related question can be treated within this framework. There, one can take $A$ to be the generator of the free streaming (semi)group.
