Let the grid be
abc
dSe
fgh
with a,b,c,d,e,f,g,h being consecutive integers not neccisarily in order.
now we have
S=2(a+c+f+h)+b+d+e+g (1)
but if we rotate then we have
bce
aSh
dfg
and then we have
S=2(b+e+g+d)+(c+a+h+f) (2)
taking (1)-(2) we end up with
a+c+f+h=b+d+e+g (3)
now let the consecutive integers be
x,x+1,x+2,x+3,x+4,x+5,x+6,x+7
by taking
a=x c=x+7 f=x+1 h=x+6 and
b=x+2 d=x+5 e=x+3 g=x+4 then we have both sides of (3)
equal to 4x+14

In general, all you need for this to work is to have the sum of the corner numbers be equal to the sum of the orthogonal numbers.