Show RxZ is a Ring with 1

Question

The problem is: $R$ is NOT a ring with "1", show that $R'=R\times \mathbb{Z} $ with Addition: $(r_1,n_1)+(r_2,n_2)=(r_1+r_2,n_1+n_2)$
and Multiplication:
$(r_1,n_1) \cdot (r_2,n_2)=(r_1r_2+n_1r_2+n_2r_1,n_1n_2)$
is a Ring with element "1", and $1=(0,1)$ (the neutral element). I have shown all the properties of a ring. That is: $(R',+)$ is a commutative group and $(R',\cdot)$ is a semigroup. I still have to show Distributive property, but that should be no problem.
What I find really confusing is to show that my neutral element $1=(0,1)$, because no matter what I do, when I multiply:
$(0,1) \cdot (r_1,n_1)=(0\cdot r_1+1\cdot r_1 + n_1\cdot 0,1\cdot n_1)= (1\cdot r_1,n_1)$ which would be no problem, but the first sentence says R is NOT a ring with "1", so I don't know if I can say that $1\cdot r_1 = r_1$. I would be grateful if someone could clarify this for me.

Answer 1

Note that $n\cdot r$ here means addition of $r$ with itself $n$ times, where $n\in\Bbb N$, so we get

$$(0,1)(r,n)=(0\cdot r+r+n\cdot 0,\ 1\cdot n) =(r, n)$$ And similarly from the right side (if $R$ is not commutative).