smc

(A) is true because multiplication is commutative. (B) is false because we have $x\star (y+z)=(x+1)(y+z+1)-1$ and $x\star y+x\star z=(x+1)(y+1)-1+(x+1)(z+1)-1$ $(x+1)(y+z+2)-2$ $(x+1)(y+z+1)-(x+1)-2$ $[(x+1)(y+z+1)-1]+x,$ which does not match with the previous expression. (C) is true because $(x-1)\star (x+1)=((x-1)+1)((x+1)+1)-1$ $x(x+2)-1$ $x^2+2x-1$ $[(x+1{)}^2-1]-1$ $(x\star x)-1$ for all $x$. (D) is true because $x\star 0=(x+1)(0+1)-1=(x+1)-1=x$ for all $x$. (E) is true because multiplication is associative (and the plus-ones inside the parentheses and the minus-ones outside cancel out). Therefore, our answer is $\boxed{x\star (y+z)=(x\star y)+(x\star z)\text{ for all real }x,y,\text{ and }z.}$, and we are done.