For any rng $R$, can we attach a unity? Let $R$ be an rng. (There may be no unity)
Then, does there always exist a ring(with unity) $A$ such that $R$ is a subrng of $A$?
 A: One can always find a ring with unity containing a ring (without unity). There are many different ways to do that ; let's look at one of them. Let $R$ denote the ring without unity and consider $A=\mathbb{Z}\times R$ with the canonical addition and the following multiplication
$$(m,a)\cdot (n,b)=(mn, na+mb+ab)$$
$A$ with these two operations is a ring and $\left(1,0_R\right)$ is a unity for that ring. And
$R\to A$ $a\to (0,a)$ is a canonical injection of $R$ in $A$
A: For future readers, the above construction can be greatly generalized.
Let $R$ be a rng (may have no unity) and $(M,+,•)$ be an $R$-module.
Define $S=\mathbb{Z}\times R$ and define operations as $(n,a)+(m,b)=(n+m,a+b)$ and $(n,a)(m,b)=(nm,ma+nb+ab)$.
Then, $R$ is a subrng of $S$ with respect to an embedding $r\mapsto (0,r)$.
Now, define $(n,a)\ast x = a•x + nx$ where $x\in M$.
Then $\ast$ indeed induces $M$ to be an $S$-module and $•$ is the restriction of $ast$ on $R\times M$.
To sum up, the following is true:

Let $R$ be an rng and $(M,+,•)$ be an $R$-module.
Then, there exists a ring $S$ and an operation $\ast:S\times M\rightarrow M$ such that $(M,+,\ast)$ is an $S$-module and $\ast\upharpoonright (R\times M)=•$.

This means that, defining module on rngs (may be without unity) is equivalent to defining module on rings (with unity).
So, by defining modules on rings with unity, one can handle both two cases.
Moreover, here is one application of the original one : "every rng is a subrng of a ring with unit"
Let $R$ be an rng and consider a polynomial rng $R[X]$.
Since $R$ may not have a unity, you cannot formally define $X$ as an element of $R[X]$.
However, by extending $R$ to a ring $S$, since $R[X]$ is a subrng of $S[X]$, you can consider $X$ as an actual object!
