FALSE: member equivalence is transitive in any pointed category with pullbacks

Work in the category of commutative rings not required to have an identity, with arbitrary ring homomorphisms. The zero ring is a zero object, and pullbacks are the usual fibre-product rings, so this category is pointed and has pullbacks. Let $A:=\mathbb{C}[t,t^{-1}],x:\mathbb{C}[t]\hookrightarrow A,y:=1_A:A\to A,z:\mathbb{C}[t^{-1}]\hookrightarrow A$ be the two localization inclusions and the identity member. The map $x$ is epic: if $\phi ,\psi :A\to R$ agree on $\mathbb{C}[t]$, then they agree on $e:=\phi (1)=\psi (1)$ and $a:=\phi (t)=\psi (t)$. Both $\phi (t^{-1})$ and $\psi (t^{-1})$ are inverses of $a$ in the commutative corner ring $eR$, so they are equal; hence $\phi =\psi$. The same argument shows that $z$ is epic. Therefore $x\equiv y$ and $y\equiv z$.

The pullback of $x$ and $z$ is the constant subring $\mathbb{C}$, because inside $\mathbb{C}[t,t^{-1}]$ one has $\mathbb{C}[t]\cap \mathbb{C}[t^{-1}]=\mathbb{C}\mathrm{.}$ Its two projections are the constant-term inclusions $p_1:\mathbb{C}\to \mathbb{C}[t],p_2:\mathbb{C}\to \mathbb{C}[t^{-1}]\mathrm{.}$

Suppose $x\equiv z$. Then there would exist a ring $T$ and epimorphisms $q:T\to \mathbb{C}[t]$ and $r:T\to \mathbb{C}[t^{-1}]$ with $xq=zr$. By the pullback property from step 1.2, there would be a map $m:T\to \mathbb{C}$ with $q=p_{1}m$ and $r=p_{2}m$. Since $q$ is epic, $p_1$ would also be epic. But $p_1$ is not epic: the evaluation maps ${\mathrm{ev}}_0,{\mathrm{ev}}_1:\mathbb{C}[t]\to \mathbb{C}$ are distinct and satisfy ${\mathrm{ev}}_0{p}_1={\mathrm{ev}}_1{p}_1=1_{\mathbb{C}}\mathrm{.}$ This contradiction shows that $x\equiv ̸z$.

Thus $\equiv$ is not transitive in this pointed category with pullbacks, and the statement is false.