NOTE
This is a follow up question to my question Existence and uniqueness for differential equations involving distributions
I am now asking that the function $z$ be continuous, otherwise my question is the same.
The Question
Consider the following DE:
$$x' = f(x) + g(x)z'$$ with initial condition: $$x(0) = x_0$$
where $f,g$ are infinitely differentiable functions of $x$ and $z'$ is the derivative (in the distributional sense) of some given continuous function z(t).
By distribution, I mean the definition given here on wikipedia.
I would like to know if there exists a unique solution $x$ (a function) such that $x$ is a solution for all t in $[-d,d]$ for some constant $d>0$.
Details
By the expression $x'$, I denote the distributional derivative of $x$ when thought of as a distribution, so $x'$ is a distribution rather than a function. Because the product of a function and a distribution is also a distribution, $g(x)z'$ is a distribution. $f$ can be interpreted as a distribution via the usual inclusion map from functions to distributions, thus we can consider the sum $f(x)+g(x)z'$ and therefore also consider the equation of distributions $x'=f(x)+g(x)z'$.
Several of the details in the form of your question turn out to be inessential, I think.
Yes, the product of a distribution by a smooth function is again a distribution. And the (distributional) derivative of a continuous function, or any locally $L^1$ function, is again a distribution.
EDIT: To be clear(er), your equation is always solvable for what you call $x$, regardless of what $f,g$ and $z$ are (just so long as the right-hand side of your equation is a distribution.
If you'll pardon me, I prefer not to call functions $x$... so I'll rewrite your equation as $u'=v$, where $u$ has replaced $x$, and the right-hand side is just denoted "$v$". That is, we do not need to use the details of what kind of distribution the right-hand side is/gives.
That is, I think we get the same conclusion from considering $u'=v$, where $v$ is a given distribution, and we are trying to solve for distribution $u$. I claim that there is always such $u$. (And we know that any two such $u$s differ by integrate-against-constant...) Certainly there is ambiguity, because differentiation kills constants (and, as one can prove, nothing else).
Existence? We're trying to solve $u'(\varphi)=v(\varphi)$, and $u'(\varphi)=-u(\varphi')$. The test functions $\varphi$ that are derivatives are (provably) those with integral $0$. So to define $u$ on all test functions it is necessary and sufficient to define it on a single fixed test function $\varphi_o$ with non-zero integral, and we can take the value $0$.
Thus, letting $1(\varphi)=\int \varphi$, we have $$ u(\varphi) \;=\; u(\varphi-1(\varphi)\cdot \varphi_o)+1(\varphi)\cdot u(\varphi_o) \;=\; u(\varphi-1(\varphi)\cdot \varphi_o)+0 $$ As already noted, the ambiguity by constants is natural.
EDIT: yes, the map that gives anti-derivatives of test functions $\varphi$ with $\int \varphi=0$ is a continuous map $T$ from that space to all test functions, given by $T\varphi(x)=\int_{-\infty}^x\varphi(t)\;dt$. Proof of this continuity does use the strict-inductive-limit definition of the topology on test functions.
EDIT: also, one can sanity-check that changing the fixed test function $\varphi_o$ indeed only changes $u$ by a constant multiple of integrate-against-$1$.
So, in summary, back to the notation of the question, yes, for all $f,g,z$, there is a distribution $x$ satisfying the differential equation. It is unique up to constant multiples of integrate-against-$1$.