Is every ring with a homomorphism from $R$ an $R$-algebra if and only if $R$ is a solid ring?

Question

Given a commutative ring $R$, is every ring with a homomorphism from $R$ an $R$-algebra if and only if $R$ is a solid ring?

A ring $R$ is said to be solid if the unique homomorphism $\mathbb{Z} \to R$ is an epimorphism in the category of rings, or equivalently, if $r \otimes 1=1 \otimes r \in R \otimes_{\mathbb{Z}} R$ for all $r \in R$.

Also, it is known that a ring homomorphism $f:R \to A$ makes $A$ into an $R$-algebra if and only if $f(R) \subseteq Z(A)$, where $Z(A)$ is the center of $A$.

The "if" direction could be proven by considering the two $R$-module structures on $A$ induced by $f$ (where $ra$ is defined to be $f(r)a$ in one structure and $af(r)$ in the other) and noting that they must be the same if $R$ is solid (as any abelian group has at most one $R$-module structure).

For the "only if" direction, let $S$ be a ring with two homomorphisms $f,g:R \to S$. Then, define a multiplication on the abelian group $R \oplus S$ with the rule $(r, s)(r', s')=(rr', f(r)s'+sg(r')+ss')$. This should then make $R \oplus S$ an $R$-algebra (with the corresponding homomorphism mapping $r \mapsto (r, 0)$) if and only if $f=g$ and $f$ itself makes $S$ an $R$-algebra.

Is that right? If so, then as every ring with a homomorphism from $R$ is assumed to be an $R$-algebra, it follows in particular that $R \oplus S$ must be an $R$-algebra, so one must have $f=g$, thus proving that $R$ is solid.

Answer 1

To update my comments: I think the statement, which I would phrase as:

In the category of (associative, unital, not necessarily commutative) rings,

$(*)$ Every homomorphism $R\rightarrow S$ has image in $Z(S)$ (i.e. makes $S$ an $R$-algebra)

is equivalent to

$(**)$ $R$ is a solid ring

is correct, and your proof works. For $(*) \implies (**)$, it might be handy to remark right away that solid rings can be characterized as the rings $R$ such that if there is any morphism $R \rightarrow S$, it is unique. And the proof could be shortened by saying first that $(*)$ implies $R$ is commutative, and instead of using your construction (which works), it might be easier to look at the ring of matrices

$$\{\pmatrix{r&s\\ 0&r} : r\in R, s \in S\}$$

where the $S$ in the upper right corner is viewed as left-$R$-module via $f$ and as right-$R$-module via $g$. That just corresponds to (but in my view, gives a good intuition for) the shorter multiplication formula

$$(r,s)(r',s') := (rr', f(r)s+s'g(r))$$

but now the idea is the same, the embedding $R \rightarrow R * Id$ gives a ring homomorphism, and by looking at multiplication with $\pmatrix{0&1\\0&0}$ a.k.a. $(0,1)$ we see the image can only be central if $f=g$.