new addition and new multiplication x ⊕ y = x + y − 1, x ⊗ y = x + y − xy on set Z, prove the set Z equipped with these 2 new operation here says a new operation addition and new operation multiplication  x ⊕ y = x + y − 1, x ⊗ y = x + y − xy on set Z,where the operations on the right hand side are ordinary addition and multiplication of integers. Prove that the set Z equipped with these two
new operations ⊕, ⊗ is a ring. Does it have a multiplicative identity?
I tried to use the addition cancellation but I do not understand why or I got the wrong understanding for this question. 
I assume the there is y and y' are 2 solution for equation  x ⊕ y = x + y − 1, therefore:
  x ⊕ y = x + y − 1
 x ⊕ y' = x + y − 1
and I get  x ⊕ y =  x ⊕ y' after the cancellation, then I get y=y', and it means the addictive is unique. So, can I say that the set Z equipped with new addiction x ⊕ y = x + y − 1? or I just go into a wrong way for solve this problem?
ask in advance, thank you

Hint $ $ Any set bijection $\,h:R'\to R\,$ serves to transport the ring structure of $\,(R,+,*,0,1),$ to $\,(R',\oplus,\otimes,0',1') \,$ by defining the operations in $\,R'\,$ to make $\,h\,$ be a ring isomorphism, i.e.
$$\ \ \  \begin{align} &x \oplus y\, =\ h^{-1}(h(x) + h(y)),\quad 0' = h^{-1}(0)\\  
 &x \otimes y\, =\, h^{-1}(h(x)\, *\, h(y)),\quad 1' = h^{-1}(1)\end{align}$$
$\begin{align}\Longrightarrow\ \ \ \ \ \ \ \  h(x\oplus y)\ =\ h(x) + h(y), & \ \ \ \ \ \ \ \ \ \ \ \ \ h(x\otimes y)\ =\ \ \ h(x) * h(y)\\[.3em]
{\rm e.g.}\ \ h(x)=1\!-\!x\!:\ \  1\! -\! (x\!+\!y\!-\!1) =\, 1\!-\!x + 1\!-\!y, & \ \ \ \ 1\!-\!(x\!+\!y\!-\!xy) = (1\!-\!x)*(1\!-\!y)\end{align}$
