No, there is no reason why the stored energy should be the same in both cases. The only requirement is that the total force supplied by each system of springs is the same, because the two systems are supporting the same load, and each is in equilibrium.
The difference in stored energy does not 'go' anywhere because these are two separate cases. You are not converting one case into the other.
If you did start with the load supported by the springs in series and converted this to the load supported by the springs in parallel, the difference in elastic energy would not in fact equal the difference in gravitational PE of the load!
The tension in each spring would have to be reduced from $W$ to $\frac12 W$ where $W$ is the weight of the load, and the extension of each reduced from $x$ to $\frac12 x$. The elastic energy stored in a spring with tension $W$ and extension $x$ is $\frac12 W x$. The total elastic energy stored in the two springs is initially (ie in series) $2\times \frac12 Wx=Wx$ and finally (ie in parallel) $2\times \frac12 (\frac12 W)(\frac12 x)=\frac14 Wx$. The decrease in elastic PE is $\frac34 Wx$.
After the change from series to parallel, the springs initially supply a total upward force of $2W$ while the weight supplies a downward force of $W$. An additional downward external force is required, starting at $W$ and decreasing to $0$, to prevent the system from gaining kinetic energy as the tension is reduced. The work done against this external force equals the KE which the load would have acquired when it reached the equilibrium position had it been released. In that case it would have overshot the equilibrium position and oscillated indefinitely.
The work done on the load (average force x distance) by the two springs is $W\times (\frac12 x)=\frac12 Wx$, which is the increase in its gravitational PE. The work done against the external force is $(\frac12 W)(\frac12 x)=\frac14 Wx$. The total work done by the two springs is $\frac34 Wx$.
To summarise : the two springs lost total elastic energy $\frac34 Wx$ of which $\frac12 Wx$ increased the gravitational PE of the load while $\frac14 Wx$ was done against an external force.
See Mass dropped on a spring.