Starting from Newton's second law, I am trying to show that energy is not conserved with an applied external force when moving a positive charge from $r_b$ to $r_a$ (see image). I'm not sure where I'm missing a minus sign? Also shouldn't the work done by the external force increase the total energy so that $E_a = W_F + E_b$ ?
You need to double check the directions of your quantities. Let's say that the $\hat r$ unit vector points to the upper right. Using this convention, the external work and the kinetic energy stay the same. The electrical force is given by $$\vec F_E=\frac{qq_0}{4\pi \epsilon_0 r^2}\hat r$$ The resulting equation in its most general form is then $$\vec F+\vec F_E=m\frac{d\vec v}{dt}=m\vec v \cdot \frac{d\vec v}{dr}$$ $$\int_{\vec r_b}^{\vec r_a} \vec F\cdot d\vec r=-\int_{\vec r_b}^{\vec r_a}\vec F_E\cdot d\vec r+\int_{\vec v_b}^{\vec v_a}m\vec v\cdot d\vec v$$ By definition, $$U(\vec r)=-\int_\infty^{\vec r}\vec F_E\cdot d\vec r$$ Hence, $$-\int_{\vec r_b}^{\vec r_a}\vec F_E\cdot d\vec r=-\int_\infty^{\vec r_b}\vec F_E\cdot d\vec r + \int_\infty^{\vec r_a} \vec F_E\cdot d\vec r=U(\vec r_b)-U(\vec r_a)$$ Therefore, $$A'=U(\vec r_a)-U(\vec r_b)+K(\vec r_a)-K(\vec r_b)$$ as you did suspect.
As you have suspected it is a matter of signs.
As an example of what you have done wrong look at your integral $\displaystyle \int_{r_{\rm b}}^{r_{\rm a}}F\,dr = F(r_{\rm a} - r_{\rm b})$.
Since $r_{\rm a} < r_{\rm b}$ this integral gives a negative value and yet the displacement of the force is in the same direction as the force!
I have annotated your derivation.
Charge $q$ is the origin of your coordinate system and with your statement that the net force is $F-F_{\rm E}$ you have defined the positive direction as in the $7$ o'clock direction.
What that means is that your starting point is at $r=-r_{\rm b}$ and your finishing point is $r=-r_{\rm a}$ as shown on the diagram above.
This assumes that you intended $r_{\rm a}$ and $r_{\rm b}$ to be positive quantities?
So now the work done by force $F$ is $\displaystyle \int_{-r_{\rm b}}^{-r_{\rm a}}F\,dr = F(r_{\rm b} - r_{\rm a})$, positive as expected.
The same change of sign will also occur for the electric potential energy terms and then you will find that the work done by force $F$ is equal to the change in kinetic energy plus the change in electric potential energy.
Perhaps you would have found it easier to choose the positive direction towards $1$ o'clock with the net force $F_{\rm E} -F$ and moving the charge from $+r_{\rm b}$ to $+r_{\rm a}$?
I have shown you how that might be done below.
System - Charge $Q_{\rm 0}$ alone
There are two external forces acting on the charge $\vec F_{\rm E}$ and $\vec F$.
You now need to consider a positive direction which can be $\hat r$ pointing roughly towards one o'clock.
This is the first place that you must be careful.
From your diagram $\vec F_{\rm E}=\dfrac{qq_0}{4\pi \epsilon_0 r^2}(\hat r)$ where $\dfrac{qq_0}{4\pi \epsilon_0 r^2}$ is a positive quantity and $\vec F = F(-\hat r)$ with $F$ chosen by you as being positive.
In moving from $\vec r_{\rm b} = r_{\rm b} (\hat r)$ to $\vec r_{\rm a}=r_{\rm a} (\hat r)$ the work done by the two forces is $\displaystyle \int_{r_{\rm b}}^{r_{\rm a}}(\vec F_{\rm E}+ \vec F)\cdot d\vec r$ where $d\vec r = dr\, \hat r$ with the sign of the component of the displacement $dr$ completely defined by the limits of integration.
This point is explained in the answer to the post Why does this line integral give the wrong sign?.
So the work done by the two external forces is
$\displaystyle \int_{r_{\rm b}}^{r_{\rm a}}(\vec F_{\rm E}+ \vec F)\cdot d\vec r=\displaystyle \int_{r_{\rm b}}^{r_{\rm a}}(F_{\rm E}(\hat r)+ F(-\hat r))\cdot dr(\hat r) =\displaystyle \int_{r_{\rm b}}^{r_{\rm a}}(F_{\rm E} -F)dr =\dfrac{qq_0}{4\pi \epsilon_0 r_{\rm a}}-\dfrac{qq_0}{4\pi \epsilon_0 r_{\rm b}}+F(r_{\rm b} - r_{\rm a})$
and this is equal to the change in the kinetic energy of the system (charge $q_0$).
From your derivation you have have chosen a different system as you cannot define a potential energy for a system consisting of a single particle/charge.
System - Charge $Q_{\rm 0}$ and charge $q$.
The assumption is then made that charge $q$ does not move / moves very little compared with charge $q_0$, ie the mass of $q$ is much greater than the mass of $q_0$.
Now there is only one external force $\vec F$ and two internal forces, the force on charge $q$ due to charge $q_0$ and its Newton third law pair, the force on charge $q_0$ due to charge $q$.
Because $q$ does not move the derivation is essentially the same and the work done by the external force is So the work done by the two external forces is $F(r_{\rm b} - r_{\rm a})$ with the change in electric potential energy being $\dfrac{qq_0}{4\pi \epsilon_0 r_{\rm a}}-\dfrac{qq_0}{4\pi \epsilon_0 r_{\rm b}} = U(r_{\rm a})-U(r_{\rm b})$ and the change in kinetic energy being $K_{\rm a}-K_{\rm b}$.
Bringing everything together gives $K_{\rm a}-K_{\rm b} = F(r_{\rm b} - r_{\rm a}) - (U(r_{\rm a})-U(r_{\rm b}))$.
If the work done by the external force, $F(r_{\rm b} - r_{\rm a})$, is greater than the increase in the electric potential energy of the system, $U(r_{\rm a})-U(r_{\rm b})$, the kinetic energy of the system increases.
