A $5\times 5$ square with $+1$ and $-1$

puzzle

Solution 1:

I agree with you that their argument is not a proof.
Here is my proof.

Consider this pattern of cells:

x . . x .
. x x . x
x . . x .
x . . x .
. x x . x

Every move toggles an even number of the cells marked x. Therefore their sum mod 4 remains constant (I actually prefer to say that the sign of their product remains constant). This means that if the initial -1 is located at one of the x-cells, then it is impossible for the moves to eliminate it - there is always an odd number of x-cells containing a -1.

This pattern works in any orientation, so every cell except for the centre one is such an x-cell. Therefore only the starting pattern with the -1 in the centre cell is solvable.

Edit:
Actually, there is simpler pattern:

x x x x x
x x x x x
. . . . .
x x x x x
x x x x x

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