John Hunter's solution is more elegant and concise; however, if you are interested also in the condition under which the point remains the neutral point and a more mathematical proof then here is a different approach:
In addition, I am not sure whether the questioner wanted the charge $q_0$ to impact the system or just as a test charge. If you wish $q_0$ to have no impact on the system then just set $q_0=0$ in the proceeding calculations.
Definitions
Because of the bulky equations, we will first define some notation:
$$\boldsymbol M\equiv\begin{bmatrix}m_0 & 0 & 0\\0 & m_1 & 0\\0 & 0 & m_2\end{bmatrix},\quad\quad\boldsymbol Q\equiv\begin{bmatrix}q_0 & 0 & 0\\0 & q_1 & 0\\0 & 0 & q_2\end{bmatrix},\quad\quad\vec{r}\equiv\begin{bmatrix}r_0\\r_1\\r_2\end{bmatrix},\quad\quad k\equiv\frac{1}{4\pi\epsilon\epsilon_0},\\\quad\\\text{and}\quad a\equiv\sqrt{q_1}+\sqrt{q_2}$$
where $m_n$, $q_n$ and $r_n$ are the mass, charge and position of the $n^\text{th}$ particle respectively. Finally, let $r_{ij}\equiv r_i-r_j$.
Equations of motion
Thus, the 1D equations of motion can concisely be written as:
$$\boldsymbol{M}\ddot{\vec{r}}=k\boldsymbol{Q}\begin{bmatrix}\frac{q_1}{r_{10}^2}+\frac{q_2}{r_{02}^2}\\\frac{q_2}{r_{21}^2}+\frac{q_0}{r_{10}^2}\\\frac{q_0}{r_{02}^2}+\frac{q_1}{r_{21}^2}\end{bmatrix}\tag{1}$$
Constraint
Now if we apply the condition that $q_0$ remains at the neutral point:
$$\left|r_{10}\right|=\frac{q_1}{a}\left|r_{21}\right|\quad\text{and}\quad\left|r_{02}\right|=\frac{q_2}{a}\left|r_{21}\right|$$
We can now use this to find the constraints on the parameters $\boldsymbol{M}$, $\boldsymbol{Q}$ and $k$ for which this is true.
Substituting these conditions into the left-hand side of (1) gives:
$$\begin{align}\boldsymbol{M}\ddot{\vec{r}}&=\frac{k}{r_{21}^2}\boldsymbol{Q}\underbrace{\begin{bmatrix}2a^2\\a^2+q_2\\a^2+q_1\end{bmatrix}}_{\equiv\vec{w}}\\\implies\ddot{\vec{r}}&=\frac{k}{r_{21}^2}\boldsymbol{M}^{-1}\boldsymbol{Q}\vec{w}\tag{2}\end{align}\\%to anyone who edits the extra line is to prevent clipping of bottom line in render$$
As $\boldsymbol{M}$ is diagonal:
$$\boldsymbol{M}^{-1}=\begin{bmatrix}\frac{1}{m_0} & 0 & 0\\0 & \frac{1}{m_1} & 0\\0 & 0 & \frac{1}{m_2}\end{bmatrix}\\%to anyone who edits the extra line is to prevent clipping of bottom of matrix in render$$
Thus equation (2) gives implies:
$$\begin{align}\ddot r_{21}&=\frac{k}{r_{21}}\left(\frac{q_2}{m_2}\left(a^2+q_1\right)-\frac{q_1}{m_1}\left(a^2+q_2\right)\right)\tag{3}\\\ddot r_{01}&=\frac{k}{r_{21}}\left(2\frac{q_0}{m_0}a^2-\frac{q_1}{m_1}\left(a^2+q_2\right)\right)\tag{4}\end{align}$$
However, we already know $\left|r_{10}\right|=\frac{q_1}{a}\left|r_{21}\right|\implies\left|\ddot r_{10}\right|=\frac{q_1}{a}\left|\ddot r_{21}\right|$
Thus, assuming the initial ordering of $r_1<r_0<r_2$ then $\ddot r_{01}=\frac{q_1}{a}\ddot r_{21}$ and substituting in equations (3) and (4) gives the constraint:
$$2\frac{q_0}{m_0}a^2-\frac{q_1}{m_1}\left(a^2+q_2\right)=\frac{q_1}{a}\left(\frac{q_2}{m_2}\left(a^2+q_1\right)-\frac{q_1}{m_1}\left(a^2+q_2\right)\right)\tag{5}$$
Similarly, if we had used $r_{20}$ we would have got:
$$\frac{q_2}{m_2}\left(a^2+q_1\right)-2\frac{q_0}{m_0}a^2=\frac{q_1}{a}\left(\frac{q_2}{m_2}\left(a^2+q_1\right)-\frac{q_1}{m_1}\left(a^2+q_2\right)\right)\tag{6}$$
Only if both (5) and (6) are satisfied then $q_0$ will remain at the neutral point. Note how both (5) and (6) only depend on charge and mass. Below I have plotted $q_2$ against $q_1$ for both equations (5) and (6) holding all other values constant for some non-zero parameters ($m_n=1$ for all $n$ and $q_0=1.7$) to demonstrate solutions do exist:
Updates
Earnshaw's theorem
I believe the point about Earnshaw's theorem is that the resultant force on all three charges cannot be simultaneously zero but it can be zero for $q_0$ as you prove in the question.
Instabilities
For attractive forces, the charges should stay in a straight line. However, for repulsive forces, small deviations from the straight line will cause the charges to move away from the 1D line and so $q_0$ will no longer remain at the neutral point regardless of whether the parameters are a solution to (5) and (6). Unless the charges are somehow restricted to the only move along the 1D line.
This can be thought of intuitively by considering whether there is a minimum (attractive case) or maximum (repulsive case) in the potential experienced by each charge perpendicular to the line. Similar to the idea of stable and unstable equilibrium; however, in that analogy we are only considering equilibrium perpendicular to the line as the system is clearly not in equilibrium parallel to the line.
