$J\trianglelefteq M_2(R)$ implies $\exists I\trianglelefteq R$ such that $J=M(I)$ 
Let $R$ be a unital commutative ring, and let $J$ be an ideal of $M_2(R)$. Then there exists an ideal $I$ in $R$ such that $J=M_2(I)$.

Proof:
Let $I:=\{a\in R: a_{1,1}\text{ is the first entry of a matrix in J} \}$. Then $I$ is not empty, since $J$ is not empty. Let $A=\begin{bmatrix}
a_{1,1} & b\\ 
c & d
\end{bmatrix}\in J$, with $b,c,d\in R$, and let $B=\begin{bmatrix}
r_1 & r_2\\ 
r_3 & r_4
\end{bmatrix}$, with $r_k\in R$, for $1\le k \le 4$. Then $AB=\begin{bmatrix}
a_{1,1}r_1+br_3 & a_{1,1}r_2 + br_4\\ 
cr_1+dr_3 & cr_2+dr_4
\end{bmatrix}\in J$. Similarly, $BA=\begin{bmatrix}
r_1a_{1,1}+r_2c & r_1b+r_2d\\ 
r_3a_{1,1}+r_4c & r_3b+r_4d
\end{bmatrix}\in J$. Thus $a_{1,1}r_1+br_3$ and $r_1a_{1,1}+r_2c$ are in $S$, where $S\le R$ (is it true that $S$ must be a subring of $R$ if $J$ is an ideal of $M_2(R)$?), such that $J = M_2(S)$. But this implies that $a_{1,1}r_1$ and $r_1a_{1,1}$ are in $S$ and $r_2c$ and $br_3$ are in $S$ (since $S$ is a ring). But then $S$ is an ideal of $R$, since $a_{1,1}\in I$ and $r_k\in R$ is arbitrary, and $a_{1,1}\in I$. Since $a_{1,1}\in I$, so $S=I$. All other entries of $R$ must also be in $S=I$, hence, $J=M_2(I)$.
But if my proof is correct, then why do we need $R$ to be unital and commutative?

Update: I see that $I$ may be a subring of $S$ in my proof, so the proof is incorrect.

 A: 
why do we need $R$ to be unital and commutative?

Firstly, we do not need $R$ to be commutative. This is perfectly true for noncommutative rings.
However, what you have written is far from a clear proof. No doubt you intended $A$ to be chosen from $J$ instead of having arbitrary entries $b,c,d$. Notice that you lose control of what they are right at that moment. You only have control over $r_1, r_2,r_3,r_4$. This indeed allows you to prove that $I$ is an ideal. Then you say

Also, $a_{1,1}\in I$, so $S=I$. 

I'm not sure what you meant by $S$, since it does not really serve a purpose. It certainly looks like (and would make sense if) you are actually thinking of $I$ again. Then

All other entries of $R$ must also be in $I$, hence, $J=M_2(I)$.

No idea what you mean. Obviously not all elements of $R$ must be in $I$. Do you mean entries of the matrix $A$ are in $I$? This is far from obvious. Why should that be? 
If the ring had an identity element then I would use permutation matrices on the left and right of $A$ to move entries of $A$ to the upper left hand corner, so that those entries would fall in $I$.  Then we could be sure that $J\subseteq M_2(I)$.
Then finally how do we know $M_2(I)\subseteq J$? If the ring had an identity then I would just take a matrix contributing $a_{11}$ and multiply it on the left and right with matrices to make it zero in all other entries. Then I could further show that this one nonzero entry could be moved into all other positions. These are all matrices contained in $J$. Then it would be clear that $M_2(I)\subseteq J$, and we would have equality.

I should have thought of this simple example a long time ago.  Let $R$ be a ring with trivial multiplication, such that $R$ has unequal additive subgroups $A$ and $A'$. Then $\begin{bmatrix}A&A\\A&A'\end{bmatrix}$ is obviously an ideal of $M_2(R)$ (because the multiplication becomes trivial in the matrix ring too.) By construction, this ideal is not of the form $M_2(I)$ for any ideal $I$. So you can see that the theorem does not hold for all rings without identity.
That's not to say the theorem fails in all rings without identity. I think everything works in rings which have local identities, for example. (This means that for every $a\in R$, there exists an idempotent $e\in R$ such that $ea=a$. One such ring is $\oplus_{i=1}^\infty F$ for a field $F$.)
