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> I'm looking for a specific construction, taking an abelian group (with designated element) $(G,+,1)$ to a commutative ring $(R,+,\cdot,1)$, where $G\subset R$ as a pointed abelian group, and which is universal in the following sense: for any commutative ring $(S,+,\cdot,1)$ and any map $f:G\rightarrow S$ preserving $+$ and $1$, there is an extension $g:R\rightarrow S$.
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The idea was to give a general construction of rings from pointed abelian groups, which in particular constructs $(\mathbb Z, +, \cdot, 1)$ from $(\mathbb Z, +, 1)$. So, while it may add new elements in some cases, it is not adding more than necessary.


The reason for adding the requirement that the groups be pointed is that there are conceivably many choices for $1$ (especially in, say, $(\mathbb Q, +)$, where every nonzero element is equally suitable).



Constructing a ring from an abelian group in a minimal way
Given an abelian group $A$ with a fixed element $e\in A$, you can construct the universal map $f$ from $A$ to a (commutative or noncommutative, as you prefer) ring $R=R(A,e)$ such $f(e)$ is the unit element in $R$. Just take the symmetric algebra $S(A)$ (if you want a commutative ring) or the tensor algebra $T(A)$ (if you allow your ring to be noncommutative) of your group $A$ considered as a module over $\mathbb Z$, and take its quotient by the ideal generated by the element $e-1$, where $e\in A$ is your given element and $1\in S(A)$ or $T(A)$ is the unit element of the symmetric or tensor algebra over $\mathbb Z$, to obtain the ring $R$. 


The natural map $f\colon A\to R$ will not be in general injective, though (i.e., not every abelian group with a fixed element can be embedded into a ring so that the fixed element becomes the unit element). E.g., if $e=0$ in $A$, then $R(A,e)=0$. If $A=\mathbb Z/4\mathbb Z$ and $e=2 \bmod 4$, then $R(A,e)=0$. Still when $A=\mathbb Z$ and $e=1$, you will get $R=\mathbb Z$ with $f\colon A\to R$ being the identity map, just as you wished.


