Prove that $I=\{x \in R : ax=0\}$ is a subring of $R$ To do so we just need to prove that I is a ring itself.  By requirements needed to be a ring,


$(I,+)$ is an abelian group: Let $x,y \in R$ then we have that $ax=0=ay$.  So $ax+ay=0=ay+ax$.
$\ast$ is associative: Let $x,y, z \in R$, which means $ax=ay=az=0$.  So $(ax*ay)az=ax(ay*az) =0$
The left and right distributive laws hold:  Let $a,b,c \in R$.  So we see that $a(b+c)=(ab)+(ac)=0$ and $(b+c)a=(ba)+(ca)=0$


Is there anything wrong with this or can anyone suggest a better way to write this?  My instructor is ridiculously picky.
To show that the structure is a sub-ring you need to show that it is closed under the relevant operations (the other restriction are inherited from the ambient ring and do not need to be checked, as mentioned in a comment).

For addition you need to show: $0 \in I$, for all $x, y \in I$ you have $x+y \in I $ and $-x \in I$ (or combied $x-y \in I$).

For multiplication you need to show: for all $x, y \in R$ you have $xy \in I$ (as mentioned in a comment). [It appears your rings do not need to have 1, otherwise you would need to assert it's in I; but this would not be possible in this case.]

So how to do this for addition. Well $a0= 0$ so $0 \in I$ that is fine. And, for $x,y \in I$ you have $ax= ay = 0$ as you said, so $0 = ax + ay = a(x+y)$ and $x + y \in I$. 
I leave the argument for $-x$ to you.
To show that the structure is a sub-ring you need to show that it is closed under the relevant operations (the other restriction are inherited from the ambient ring and do not need to be checked, as mentioned in a comment).

For addition you need to show: $0 \in I$, for all $x, y \in I$ you have $x+y \in I $ and $-x \in I$ (or combied $x-y \in I$).

For multiplication you need to show: for all $x, y \in R$ you have $xy \in I$ (as mentioned in a comment). [It appears your rings do not need to have 1, otherwise you would need to assert it's in I; but this would not be possible in this case.]

So how to do this for addition. Well $a0= 0$ so $0 \in I$ that is fine. And, for $x,y \in I$ you have $ax= ay = 0$ as you said, so $0 = ax + ay = a(x+y)$ and $x + y \in I$. 
I leave the argument for $-x$ to you.