Given a morphism of commutative rings $f:A\to B$, we define its epicenter, or dominion, by the set $$E=\{b\in B \;|\; b\otimes_A 1 = 1\otimes_A b\}.$$Note that if $u,v:B\to C$ are morphisms of commutative rings such that $u\circ f =v\circ f$, then for all $b\in E$, $u(b)=v(b)$. In fact, it is known that $f$ is an epimorphism if and only if $E=B$.
Now assume $E\subsetneq B$. Since $f(A)\subset E$, we can consider the corestriction $f\colon A\to E$. Is it an epimorphism? It seems likely and we already know it is when we can extend morphisms $E\to C$ to morphisms $B\to C$.
I believe the answer is "no". In general (other categories), this process is not guaranteed to yield an epimorphism. For example, there are categories where there are nontrivial dominions (the dominion properly contains the image), but all epimorphisms are surjective, so that this process need not yield an epimorphism. One such example is the variety of nilpotent groups of class at most $2$. See References 1 and 2.
Unless I messed up (and I will welcome any corrections), here is an example where you do not get an epimorphism. There may be simpler examples, but I got this one by thinking about a "universal counterexample" based on the Silvert-Mazet-Isbell Zigzag Lemma (see below).
(I will use "dominion" slightly different, because that is how I know the term: given a subring $D$ of $B$, the dominion of $D$ in $B$ is $$E=\{b\in B\mid \text{for all rings }C,\text{ for all }f,g\colon B\to C,\text{if }f|_C=g|_C\text{ then }f(b)=g(b)\},$$ where $C$ ranges over all commutative unital rings. In the case of commutative rings with unity, one can prove that this is equal to the set of all $b\in B$ such that $b\otimes_D 1 = 1\otimes_D b$, as you describe. The connection with your definition is that what you call the dominion of $f$ is what I call the dominion of $f(A)$ in $B$. But because $f$ can be factored into a surjection followed by an inclusion, and the surjection is an epimorphism, the function is an epimorphism if and only if the dominion of $f(A)$ in $B$ is all of $B$. So I will restrict myself to the situation where $A$ is a subring of $B$ and our morphism is just the canonical inclusion $A\hookrightarrow B$.)
The Silver-Mazet-Isbell Zigzag Lemma (see Reference 3) gives an explicit description of the dominion of $A$ in $B$: it consists of the elements of $B$ that can be written as $XYZ$, where $X$ is a row vector, $Z$ is a column vector, $Y$ is a matrix, $X$ and $Z$ have coefficients in $B$, and $XY$, $Y$, and $YZ$ have coefficients in $A$. It is easy to verify all such elements lie in the dominion; the "meat" of the lemma is proving the converse. Note also that the dominion operator is idempotent: that is, the dominion of the dominion of $A$ is just the dominion of $A$.
Let $B=\mathbb{Z}[x,y,z]$. Let $A$ be the subring $A=\mathbb{Z}[xy,y,yz]$, and take $f$ to be the embedding. Note that $A$ consists of the polynomials in which, in every monomial, the power of $y$ is at least the sum of the powers of $x$ and $z$.
I claim that the dominion of $A$ in $B$ is precisely: $$E = \{r+p(x,y,z)\in B\mid r\in\mathbb{Z}, p(x,0,z)=0\}=\mathbb{Z}+yB,$$ that is, it consists of all polynomials of the form $a+p(x,y,z)$, where $a$ is constant and $p(x,y,z)$ is a multiple by $y$. Note that this is strictly larger than $A$, since it contains, for example, $xyz$.
To verify all such elements lie in the dominion of $A$, it suffices to show that if $x^ay^bz^c$ lies in the dominion and $b\gt 0$, then $x^{a+1}y^bz^c$, $x^ay^{b+1}z^c$, and $x^{a}y^bz^{c+1}$ also lie in the dominion.
Indeed, if $f,g\colon B\to C$ are two ring homomorphisms, $f|_A=g|_A$, and $f(x^ay^bz^c)=g(x^ay^bz^c)$, then:
Since all monomials that are multiples of $y$ lie in the dominion of $A$, it follows that $E$ is contained in the dominion.
To prove the other inclusion, let $p(x,y,z)\in B$, $p(x,y,z)\notin E$. Now consider the maps $f,g\colon B\to B$ determined by $$\begin{align*} f:\ x&\mapsto x & y&\mapsto 0 & z&\mapsto z\\ g:\ x&\mapsto 0 & y&\mapsto 0 & z&\mapsto 0 \end{align*}$$ Because the two maps send $y$ to $0$, they agree on $A$, and the value on any element of $A$ is an integer. But because $p(x,y,z)$ is not in $E$, it must be the case that $p(x,0,z)$ is not constant (otherwise $p$ is of the form $a+yq(x,y,z)$, and hence lies in $E$). Now, $g(p)=p(0,0,0)=0$; but $f(p)=p(x,0,z)\neq 0$. Thus, $f(p)\neq g(p)$, so $p(x,y,z)$ is not in the dominion. Thus, the dominion is contained in $E$, and we have equality.
So now we consider the embedding $A\subseteq E$. I claim that $xyz$ does not lie in the dominion of $A$ in $E$.
Let $$ C = \{a+b\epsilon\mid a,b\in\mathbb{Z}, \epsilon^2=0\}$$ be the ring of dual numbers. Write elements of $E$ as $$ a_0 + a_1y + a_2xy + a_3yz + a_4xyz + h(x,y,z)$$ every monomial of $h(x,y,z)$ is divisible by $x^2y$, $y^2$, or $yz^2$, and $h(0,0,0)=0$. Define $f,g\colon E\to C$ by $$\begin{align*} f\Big(a_0+a_1y+a_2xy+a_3yz+a_4xyz+h(x,y,z)\Bigr)&= a_0 + (a_1+a_2+a_3+a_4)\epsilon,\\ g\Bigl(a_0+a_1y+a_2xy+a_3yz+a_4xyz + h(x,y,z)\Bigr) &= a_0 + (a_1+a_2+a_3)\epsilon. \end{align*}$$ The maps are clearly additive.
To verify that they are multiplicative, note that the product any two of $y$, $xy$, $yz$, and $xyz$ is divisible by $x^2y$, $y^2$, or $yz^2$, so if $$\begin{align*} p &= a_0 + a_1 y + a_2xy + a_3yz + a_4xyz + h_1(x,y,z)\\ q &= b_0 + b_1 y + b_2xy + b_3yz + b_4xyz + h_2(x,y,z) \end{align*}$$ then their product can be written as $$\begin{align*} a_0b_0 + (a_0b_1+a_1b_0)y &+ (a_0b_2+a_2b_0)xy + (a_0b_3+a_3b_0)yz\\ &\quad + (a_0b_4+a_4b_0)xyz + H(x,y,z), \end{align*}$$ with every term of $H(x,y,z)$ divisible by $x^2y$, $y^2$, or $yz^2$. Thus, $$\begin{align*} f(pq) &= a_0b_0 + \Bigl(a_0(b_1+b_2+b_3+b_4)+b_0(a_1+a_2+a_3+a_4)\Bigr)\epsilon\\ f(p)f(q) &= \Bigl(a_0 + (a_1+a_2+a_3+a_4)\epsilon\Bigr)\Bigl(b_0+(b_1+b_2+b_3+b_4)\epsilon\Bigr)\\ &= a_0b_0 + a_0(b_1+b_2+b_3+b_4)\epsilon + b_0(a_1+a_2+a_3+a_4)\epsilon = f(pq).\\ \end{align*}$$ and similarly for $g(pq)=g(p)g(q)$, except that in both expressions $a_4$ and $b_4$ are omitted.
Since $f|_A = g|_A$, but $f(xyz) = \epsilon\neq g(xyz)$, it follows that $xyz$ does not lie in the dominion of $A$ in $E$.
Thus, the correstriction of the embedding $\mathbb{Z}[xy,y,yz]\hookrightarrow \mathbb{Z}[x,y,z]$ to the dominion is not an epimorphism.
Note that neither $f$ nor $g$ can be extended to a morphism $\mathbb{Z}[x,y,z]\to C$. The image of $x$ would have to be of the form $1+a\epsilon$ (so that $xy$ maps to $\epsilon$); but then the image of $x^2y$ would be $(1+a\epsilon)^2\epsilon = \epsilon$, whereas both $f$ and $g$ map it to $0$.
References