A subring may have a different identity,why?

First, one must be careful in how one defines rings.

From the point of view of universal algebra, there are two related structures: rings, and rings with unity.

Rings have four operations: a binary operation $+$, a binary operation $\cdot$, a unary operation $-$ (additive inverse), and a nullary operation $0$ (additive neutral element), and satisfies a bunch of identities (which make $(R,+,-,0)$ into an abelian group, $(R,\cdot)$ into a semigroup, and where $\cdot$ distributes over $+$ on both sides). Substructures are required to be closed under the operations, and morphisms are required to respect the operations.

Rings with unity have five operations: in addition to the four operations and identities mentioned above, there is a second nullary operation, $1$, which makes $(R,\cdot,1)$ into a monoid. Substructures are required to respect all operations, so subrings-of-rings-with-unity are required to have the same unity as the original ring (just like a submonoid is required to have the same identity as the original monoid). And morphisms are required to respect all operations, so that morphisms are required to be unital (send $1$ to $1$). For example, this is the convention in Lam’s First Course in Noncommutative Rings.

If you do not require rings to have an identity, then the existence of one is more happenstance than structure; this is much like the way in which a subgroup of a group may happen to be abelian, even if you do not require your groups to be abelian; or how a semigroup may happen to have an identity, even though you do not require things to be monoids.

When that happens, the situation you are seeing is the situation that occurs with semigroups and monoids. A semigroup may happen to be a monoid (have a multiplicative identity), but not every subsemigroup will necessarily have an identity, or even if it does it need not be the same as the identity of the original semigroup.

It may seem weird because we are used to groups, not semigroups. But this kind of behavior happens in semigroups all the time. In fact, given any semigroup $S$, one may extend it to a larger semigroup by adding an element $1$ that acts as an identity, even if $S$ already had one! So you can have an infinite increasing sequence of semigroups $S_0\subseteq S_1\subseteq S_2\subseteq\cdots$ where $S_i$ is a proper subsemigroup of $S_{i+1}$, and $S_i$ has an identity for each $i\gt 0$, and the identity of $S_i$ is different from the identity in $S_{i+1}$. And because rings are only required to be semigroups under multiplication, this tells you that there is no reason why this kind of behavior will not occur in rings as well (and in fact, it does... given any ring $R$, there is a construction, called the Dorroh extension, that embeds $R$ into a ring with identity that is strictly larger than $R$, and where $R$ is a proper ideal of the new ring; even if $R$ already had an identity).

So, yeah, it’s weird when one is used to groups, but it shouldn’t be. Or at least, it will seem less weird with some experience.

I really like Arturo Magidin's excellent answer, but I think there is another interesting perspective on what's going on here.

Idempotents

What's the relevant key difference between say groups and monoids, or groups and rings, or integral domains and rings, that explains why the first has no "nontrivial" "unitless subobjects" which have an identity different from the large object identity? I'm using "unitless subobject" to mean a nonempty subset closed under all operations except multiplicative identity. I'll explain what I mean by nontrivial later.

From my perspective, the answer is existence of "nontrivial" idempotents.

So what's an idempotent?

An idempotent element $e$ is an element such that $e^2=e$.

Example: In any monoid or ring or group, $1^2=1$, so the identity is an idempotent. In a ring $0^2=0$ is also an idempotent. These are the "trivial" idempotents, in the sense that these idempotents are forced to exist by the axioms of the object in question.

So why are idempotents relevant?

Well, if $S\subseteq R$ is a subring of a ring $R$ with a different unit $1_S\ne 1_R$, then $1_S^2=1_S$, so $1_S$ is an idempotent in $R$. Similarly for $N\subseteq M$ a subsemigroup of a monoid or group which has its own identity $1_N\ne 1_M$, we must have $1_N^2=1_N$.

Conversely, if $e\in R$ is an idempotent, you can check that $eRe = \{ere : r\in R\}$ is a subring of $R$ with unit $e$. (Similarly, in the monoid case, if $e$ is an idempotent, then $eMe$ is a subsemigroup with unit $e$).

Note that if $S$ has identity $1_S$, then for all $s\in S$, $1_Ss1_S=s\in 1_SR1_S$, so for any idempotent $e$, $eRe$ is essentially the largest subring on which the idempotent $e$ behaves like a unit element. (Same for monoids)

In the case of the trivial idempotents, we get the subrings $1R1=R$ and $0R0=0$.

If there are no nontrivial idempotents, then the object can't have any interesting "unitless subobjects" which have units different than the unit of the whole object.

In the case of groups, the equation $e^2=e$ implies $e=1$, since $e$ is invertible, and for rings $e^2=e$ implies $e(e-1)=0$, which if $R$ is an integral domain implies $e=0$ or $e=1$.

In your particular example, $1_S = \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$ and $S=1_SR1_S$, so this is an example of the general pattern.

Rings, idempotents, and projections

For rings we actually have more interesting behaviour. A common source of idempotents is the following situation:

Perhaps our ring $R$ is the product of two rings $S$ and $T$, so $R=S\times T$. The identity of $R$ is $(1,1)$, and we have two idempotents, the identity of $S$, $(1,0)$, and the identity of $T$, $(0,1)$, and $S=(1,0)R(1,0)$ and $T=(0,1)R(0,1)$.

We might ask, when is the converse true? I.e., if $e$ is an idempotent of $R$, when does $e$ arise as the image of $(1,0)$ under some isomorphism $R\cong S\times T$ for some rings $S$ and $T$.

Well, first notice that there are two necessary conditions, $e$ is central, since $(1,0)$ would be in $S\times T$, and that $1-e$ is also a central idempotent.

You can check that $1-e$ is always an idempotent, and if $e$ is central, then so is $1-e$.

In fact this is sufficient as well.

If $e$ is a central idempotent, then $R\cong (Re)\times (R(1-e))$, via the map $r\mapsto (re,r-re)$, which has inverse $(ae,b(1-e))\mapsto ae+b(1-e)$.

Proof sketch/notes:

Checking that these maps are inverse bijections is straightforward and doesn't use centrality. Additivity of $r\mapsto (re,r-re)$ also doesn't require centrality of $e$, so in general we always get a decomposition of $R$ as the direct sum of left $R$-ideals $Re\oplus R(1-e)$, which can be useful. However, when $e$ is central, then $$(re,r(1-e))(se,s(1-e)) = (rese,r(1-e)s(1-e)) = (rsee,rs(1-e)(1-e))=(rse,rs(1-e)),$$ which is what's required to show that the homomorphism $R\to Re\times R(1-e)$ is multiplicative.