Prove that $(H,+,\circ)$ is unital commutative ring. Suppose that $(H,+,.)$ is a unital commutative ring. Now for all $x,y \in H$ define $$x \circ y= xy+yx.$$
With this operation we have another ring which is also commutative. $(H,+,\circ)$
Now I have to show that this ring is also unital under $\circ$ operation. 
My observation:  
I have to find $e\in H$  such that $x\circ e=x$
$xe+ex=x$ which gives $e=1/2$ but we don't know the elements of $H$ so we can't say $1/2\in H$, can we? 
Please guide me. Thank you. 
 A: Since $H$ is commutative, $x\circ y=xy+yx=2xy$. This new multiplication is obviously commutative, but need not have a neutral element. For instance, if $H$ has characteristic $2$, we have $x\circ y=0$ for every $x,y\in H$.
Is the operation associative? $(x\circ y)\circ z=(2xy)\circ z=4xyz$ and $x\circ(y\circ z)=2x(y\circ z)=2x(2yz)=4xyz$.
Distributivity is obvious.
Note. If $n$ is an integer and $x\in H$, $nx$ is the standard multiple.
Under what condition does the circle operation have a neutral element $e$? We need it satisfies $2ex=x$, for every $x$, in particular for $\mathbf{1}$ (the neutral element of $H$), so $2e=\mathbf{1}$. This condition is also easily seen to be sufficient: indeed, if $2e=\mathbf{1}$, then $e\circ x=2ex=\mathbf{1}x=x$.
Thus the ring $(H,+\circ)$ is unital if and only if $2\mathbf{1}=\mathbf{1}+\mathbf{1}$ is invertible in $(H,+,\cdot)$. This is granted if the characteristic of $H$ is finite and coprime with $2$. It is false whenever the characteristic is divisible by $2$: if the characteristic is $2k$, then $k\mathbf{1}=(k\mathbf{1})\circ e=2ke=0$ gives a contradiction.
If the characteristic is $0$, it may happen or not: the identity doesn't exist for $H=\mathbb{Z}$, it obviously exists for $H=\mathbb{Q}$.
A: As is pointed out in the comments, the construction which replaces $xy$ with
$x \circ y = xy + yx \tag 1$
will leave the ring $\Bbb Z$ unitless under the multiplication operation "$\circ$";
first we note that since $H$ is commutative,
$x \circ y = xy + yx = xy + xy = 2xy; \tag 2$
then if 
$\exists e \in \Bbb Z, \; x \circ e = e \circ x = x, \tag 3$
we would also have, via (2),
$x \circ e = 2xe; \tag 4$
but then from (3),
$2xe = x \Longrightarrow 2xe - x = 0 \Longrightarrow x(2e - 1) = 0; \tag 5$
since $\Bbb Z$ is an integral domain, if $x \ne 0$ we obtain
$2e - 1 = 0, \tag 6$
which has no solution in $\Bbb Z$.
Will this construction work in $\Bbb Q$?  In this case, we still find that (2) binds; therefore so do (4)-(6), and taking
$e = \dfrac{1}{2}, \tag 7$
we see that
$x \circ e = x \circ \dfrac{1}{2} = 2 \dfrac{1}{2} x = x, \tag 8$
so apparently $e = 1/2$ is indeed the unit of $(H, +, \circ) = (\Bbb Q, +, \circ)$.  As a check, we observe that
$e \circ e = 2 \left ( \dfrac{1}{2} \right )^2 = 2 \dfrac{1}{4} = \dfrac{1}{2} = e, \tag 9$
so $e$ is idempotent with respect to $\circ$, as should be true of any unit.
Nota Bene:  Caveat Emptor!!!  I have confined my remarks here to the specific questions surrounding $e$; I have granted our OP Hitman's (perhaps tacit) assertion that indeed $(H, +, \circ)$ does indeed satisfy all the ring axioms, such as associativity, distributivity, etc, without checking them myself. End of Note.
