Unity of Subring is not necessarily Unity of Ring

Theorem
Let $\struct {S, +, \circ}$ be a ring with unity whose unity is $1_S$.

Let $\struct {T, + \circ}$ be a subbring of $\struct {S, + \circ}$ whose unity is $1_T$.

Then it is not necessarily the case that $1_T = 1_S$.

Proof
Let $\struct {S, + \times}$ be the ring formed by the set of order $2$ square matrices over the real numbers $R$ under (conventional) matrix addition and (conventional) matrix multiplication.

Let $T$ be the subset of $S$ consisting of the matrices of the form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ for $x \in \R$.

We have that:
 * $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} -y & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} x - y & 0 \\ 0 & 0 \end{bmatrix}$

and so $T$ is closed under subtraction.

From Matrices of the Form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ we have that $\struct {T, \times}$ is a subsemigroup of $\struct {S, \times}$.

It follows from the Subring Test that $\struct {T, + \circ}$ be a subbring of $\struct {S, + \circ}$.

From (some result somewhere) $\struct {S, \times}$ has an identity $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ which is not in $\struct {T, \times}$.

However, note that:

demonstrating that $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ is the identity of $\struct {T, \times}$.

The result follows.

Also see

 * Identity of Submonoid is not necessarily Identity of Monoid