Ring homomorphisms that generate the unit ideal We generally do geometry over commutative rings by insisting that all rings have a $1$, and that morphisms preserve $1$.
This corresponds, roughly, to the following geometric property: if $f:X\to Y$ is a morphism of affine schemes, then a non-vanishing section $\sigma$ on $Y$ pulls back to a non-vanishing section $f^\ast \sigma$ on $X$.
But it seems to me that we could weaken this a little bit, and just insist that a ring morphism $\varphi: R\to S$ have the following property: $\varphi(R)$ generates $S$ as an ideal.
As quid and Eric Wofsey have pointed out: We can show, for such morphisms, that if $R$ and $S$ have $1$, then $\varphi(1) = 1$, so this is equivalent to the usual definition for rings with unity, but it may give something interesting for rings without unity.
This definition seems to have a few other nice properties: for example, these morphisms are closed under composition.
My question is quite simple: is this a well-studied definition, and does anyone have references along these lines?  If it hasn't been studied, could someone explain what flaws prevent this from being a useful definition?

An example when $S$ has $1$ but $R$ does not:
Let $R$ be the sub-ring (without unity) of $\mathbb{Z}[X,Y]$ consisting of those polynomials with zero constant term.  There is a ring morphism $\varphi:R\to \mathbb{Z}$ sending $X$ to $2$ and $Y$ to $3$.  Then $\varphi(R)$ generates $\mathbb{Z}$ in a non-trivial fashion.

Here is a proof that if $\varphi: R\to S$ is such a morphism, and $R$ has $1$, then $S$ has $1$:
Let $a=f(1)$, and take any $s\in S$.  Since $t$ generates $S$ as an ideal, there is some $t\in S$ such that $at = s$, and likewise there is some $u\in S$ such that $au=t$.  But then $t = au = a^2 u = a(au) = at = s$, so $t=s$, therefore $s=at=as$.  We conclude that $a$ is a multiplicative identity in $S$.

One has $\varphi(1)= \varphi(1^2)= \varphi(1)^2$ and thus $\varphi(1)$ is idempotent. 
In addition you observed $\varphi(1)$ is invertible if there is a $1 \in S$. 
Thus $\varphi(1)$ is a unit and idempotent, which means it is $1$. 
Thus, at least in the presence of $1$, this does not give anything new. 
Also if there is no $1$ the condition that $\varphi(1)$ is idempotent and still $\varphi(R)$ generates as ideal seems very restrictive. Is there any non-trivial example?  
