This equation is said to "reduce to quadratures": you can essentially solve it exactly, in the sense that you get your solution as a well-defined integral. This integral is perfectly fine as a function, and it can be used if you so wish to calculate the solution numerically. Unfortunately, while you can do this integral exactly, the final step requires an inversion step which cannot be done using elementary functions.
To do this, one uses the Work-Energy Theorem, which is actually much more useful than you've realized so far. To apply it, simply multiply both sides of the equation by $x'(t)$ and integrate:
$$
\begin{align}
mx''(t)&=\frac{kQq}{(x(t))^2}
\\\Rightarrow
mx''(t)x'(t)&=\frac{kQq}{(x(t))^2}x'(t)
\\\Rightarrow
m\int x''(t)x'(t)dt&=\int\frac{kQq}{(x(t))^2}x'(t)dt
\\\Rightarrow
\frac12m(x'(t))^2&=\int\frac{kQq}{x^2}dx=\frac{kQq}{x(t)}+E,
\end{align}
$$
so that therefore
$$
x'(t)=\sqrt{\frac{2}{m}}\sqrt{\frac{kQq}{x(t)}+E}
$$
and by rearranging and integrating
$$
\int \frac{x'(t)dt}{\sqrt{\frac{kQq}{x(t)}+E}}=\int\sqrt{\frac{2}{m}}dt,
$$
and finally
$$
t-t_0=\sqrt{\frac{m}{2}}\int_{x(t_0)}^{x(t)} \frac{dx}{\sqrt{\frac{kQq}{x}+E}}.
$$
This integral can indeed be done exactly. Mathematica tells me that it equals
$$
t-t_0=\sqrt{\frac{m}{2}}\left[\sqrt{\frac{kQq}{x}+E}\frac{x}{E}-\frac{kQq}{2E^{3/2}}\ln\left(kQq+2Ex+2x\sqrt{E}\sqrt{\frac{kQq}{x}+E}\right)\right]_{x(t_0)}^{x(t)},
$$
although depending on the sign of $E$ you may be able to achieve simpler expressions in terms of inverse hyperbolic functions. (The case where $E=0$ is also simple to handle.) The problem after that, though, is that this function is essentially impossible to invert analytically. Your solution is "the inverse of the function above", which definitely exists and is continuous and differentiable, but it does not have an analytical expression in terms of elementary functions.
