If $g(x)$ is odd with respect to $x$, then is $$\int_{x-ct}^{x+ct}g(s)ds$$
also odd with respect to $x$? I tried by fundamental theorem of calculus that $$-\int_{x-ct}^{x+ct}g(s)ds=-(G(x+ct)-G(x-ct))$$
where $G'=g$ and I think $G$ is actually even. I'm trying to of course show that $$-\int_{x-ct}^{x+ct}g(s)ds=\int_{-x-ct}^{-x+ct}g(s)ds$$
but I can't make progress.
Your approach at the end is the right one, FTC just complicates things and would only apply to continuous functions anyway.
Call $A(x): = \int_{x-ct}^{x+ct} g(s)ds$. As you say, you want to show $A(-x) = -A(x).$ So start with
$$-A(x) = - \int_{x-ct}^{x+ct} g(s)ds $$
then pull the $-$ into the integral, then into $g$ (via $g$ being odd). You get
$$ \int_{x-ct}^{x+ct} g(-s)ds$$
Now see what happens when you try the substitution $u=-s$. Can you show (maybe in two steps) that indeed you get:
$$\stackrel{!}= \int_{-x-ct}^{-x+ct} g(u)du = A(-x)$$