What you've discovered is the reason for oscillations!
Let's call the position where the string is not stretched $x=0$. Suppose we just hold the ball at this position. The totaly energy of the ball at this point is $0$ in our conventions.
Now, we leave it and let it go down. The forces balance at $x_{eq} = - \tfrac{mg}{k}$. And, as you correctly pointed out, the potential energy of the ball at this point is
$$V_{eq} = mgx_{eq} + \tfrac{1}{2} k x_{eq}^2 = - \tfrac{1}{2} k x_{eq}^2.$$
And, your question: why is this value not $0$?
The answer is: the rest of the energy is in the kinetic energy at this point. The speed at this point will have the correct value such that $\tfrac{1}{2} m v_{eq}^2 = \tfrac{1}{2} k x_{eq}^2$, and therefore the ball continues going down despite the forces being balanced.
Now, the potential for confusion is that the ball will eventually come to rest, so where's the enegy then. The answer here is dissipation: there has to be some force damping the oscillations, otherwise they'll never stop and the energy will always be $0$.
There's an alternative picture where you never let the ball gain any velocity by putting a stage below it and slowly moving the stage down. I don't have a copletely clear understanding of this, but I think the stage eats up the difference in energy in some way. A slightly simpler case is when you move the stage not continuously but in tiny discrete steps; then clearly at each step the stage is eating up enough energy to damp the oscillations. I don't understand the limit of this discrete process as the step size goes to $0$, but that's all that needs to be understood.
Aside: the fact that the total energy at the end when the ball has come to rest at the equilibrium point is negative is, in some guise, the virial theorem of classical mechanics. I haven't unpacked exactly how, but let me flesh out the analogy in my mind a bit; if anyone can say it more precisely, pleas edit this answer.
The classical example of the virial theorem is a planet in orbit around a star. In this case, the gravitational potential energy is $-2$ times the kinetic energy of rotation. If you concentrate only on the radial direction, it looks an awful lot like there's a centrifugal force pushing away from the star and gravitational energy is pushing towards it. The analogy is centrifugal force $\leftrightarrow$ $mgx$ and gravitational force $\leftrightarrow \tfrac{1}{2} k x^2$, counter-intuitive as that is.
Step 1 on the way to making it precise is to notice that the gravitational force $\tfrac{G M m}{r^2}$ can be expanded for small displacements --- so that $r=r_0 + \delta r$ --- as $\tfrac{G M m}{r_0^2} (1 - 2 \delta r)$, giving a linear force. I don't know how to think of the centrifugal force though.
Acknowledgements for discussion: user128785 and another friend who's not on stakc exchange.
EDIT: Turns out Timaeus said the same thing first. Apologies for repeat.