First, a counterexample: for $-a_1=a_2=1,g=1,f(x)=1+x^2$, the maximal interval $I_{(0,0)}$ is $[-\pi/4,\pi/4]$, not open!
So I am assuming $f$ is actually defined on $(a_1,a_2)$. The “proof” below isn’t complete, some detail is omitted.
First, it is easy to show that if $F$ is a given antiderivative of $1/f$ (so $F$ monotonous), $G$ some antiderivative of $g$, $x$ is a solution of the differential equation iff $x=F^{-1}(G(t)+c)$ for some constant $c$.
As a consequence, $(t,t_0,x_0) \in D$ iff $t_1 < t,t_0<t_2$ and $a_1<x_0<a_2$ and there exists some $c$ such that $F^{-1}(G(t_0)+c)=x_0$ and $G(t)+c$ is in the domain of $F^{-1}$ (which we denote as $(b_1,b_2)$). Note that in this case, $c=F(x_0)-G(t_0)$.
Thus, $(t,t_0,x_0) \in D$ iff $t,t_0 \in (t_1,t_2)$, $x_0 \in (x_1,x_2)$ and $G(t)-G(t_0)+F(x_0) \in (b_1,b_2)$. This is clearly an “open condition”.
For the second part, just note from the above that $\varphi_{(t_0,x_0)}(t)=F^{-1}(G(t)-G(t_0)+F(x_0))$, hence the global flow is clearly continuous.
As you did in the periodic case or similarly over the time interval $[0,1]$, you can construct a linear bound over any time interval $[-N,N]$. The coefficients may grow with $N$, but by the continuity you know that the bounds exist. Uniqueness tells us that the restriction of the $N+1$ solution to the $N$ interval has to coincide with the $N$ solution.
See also https://math.stackexchange.com/questions/347599/questions-about-the-picard-lindelöf-theorem and the links there for all-at-once proof ideas.
Best Answer
Existence of solution follows from the continuity of $f$. Suppose that $x_1(t)$ and $x_2(t)$ are solutions and let $h(t)=(x_2(t)-x_1(t))^2$. Then $$ h'(t)=2(x_2(t)-x_1(t))(x_2'(t)-x_1'(t))=2(x_2(t)-x_1(t))\bigl(f(t,x_2(t))-f(t,x_1(t))\bigr)\le0. $$ Then $h$ is decreasing, non-negative and $h(t_0)=0$. The only possibility is that $h(t)=0$ for $t\in[t_0,t_1]$.