Ring which satisfies that if $S$ is subring of $R$ then $R^\times\cap S\ne S^\times$ Let $R$ be a unital ring and $S\le R$ be a unital subring with $1_S = 1_R$.

Then it can be shown that $S^\times\subset R^\times\cap S$, where $R^\times$ and $S^\times$ are the sets of units of respective rings. (Statement *)
There can be found a ring $R$ and its subring $S$ such that  $R^\times\cap S\ne S^\times$. That is, $\exists a\in R^\times\cap S$ such that $a\not\in S^\times$. (Statement **)

I have a couple of questions, as I'm feeling quite confused:
(1) Is it possible that $S\le R$ and yet $1_S\ne 1_R$?
(2) I can't think of an example of a ring and a subring that satisfy the Statement **. Can someone please give an example?

For (1), consider the (unital) ring $\mathbb{R}^2$ where addition and multiplication are calculated coordinate-wise, i.e.
$$(a,b)+(c,d) = (a+c,b+d)$$
$$(a,b)\cdot (c,d) = (a\cdot c,b\cdot d)$$
Then the subset $\mathbb{R}\times \{0\} = \{(a,0) \mid a\in\mathbb{R}\}$.  In this case, $\mathbb{R}\times \{0\}$ is a subring of $\mathbb{R}^2$ since it is closed under addition and multiplication and share the same additive identity. However, it is also unital with unit $(1,0)\neq (1,1)$, so it isn't a unital subring of $\mathbb{R}^2$.
For (2), consider the ring $\mathbb{R}$ with the usual operations.  It has unital subring $\mathbb{Z}$.  Then $\mathbb{R}^\times = \mathbb{R}\setminus\{0\}$, and clearly $\mathbb{R}^\times \cap \mathbb{Z} = \mathbb{Z}\setminus \{0\}$, though $\mathbb{Z}^\times = \{\pm 1\}$.
