I claim that a such category exist: Let $\mathcal{C}$ any category, build a full immersion $\mathcal{C}\subset \mathcal{C'} $ adding to $\mathcal{C}$ the new objects: $\coprod_f d_0(f), \ \coprod_f d_1(f) $ and the no-identity new arrows: $\coprod_f f: \coprod_f d_0(f) \to \coprod_f d_1(f)$ and for any $f: X\to Y$ the arrows $\epsilon^0_f: X\to \coprod_f d_0(f)$ (think as a f-coprojection of $X=d_0(f)$) and $\epsilon^1_f: X\to \coprod_f d_1(f)$. Then the $\mathcal{C'}$ arrows are the $\mathcal{C}$ arrow more the following:
$\coprod_ff\star \epsilon^0_g\star h $ , $ \star \epsilon^0_g\star h $, $ \epsilon^1_g\star h $ ($h\in \mathcal{C}\downarrow d_0(g) $) with obvious composition (here "$\star$" is a "free" composition, and composing morphisms of $\mathcal{C}$ by original compositionlaw " $\circ$ " whenever possible).
Then in $\mathcal{C'}$ we consider the congruence : $\coprod_ff\star \epsilon^0_f\star g \sim \epsilon^1_f\star (f\circ g) $
And let $\mathcal{C''}$ the quotient category, still we have a full immersion $\mathcal{C}\subset \mathcal{C''} $
And in $\mathcal{C''}$ we have the commutative diagram:
1] $$\require{AMScd} \begin{CD} X @>{\epsilon^0_f}>> \coprod_fd_0(f)\\ @V{f}VV @VV{\coprod_ff}V \\ Y @>{\epsilon^1_f}>> \coprod_fd_1(f) \end{CD}$$
i.e. $\coprod_ff\circ \epsilon^0_f = \epsilon^1_f \circ f$, but the follow isn't commutative (commutative only if $f=g=h$):
2] $$\require{AMScd} \begin{CD} X @>{\epsilon^0_g}>> \coprod_fd_0(f)\\ @V{h}VV @VV{\coprod_ff}V \\ Y @>{\epsilon^1_f}>> \coprod_fd_1(f) \end{CD}$$
i.e. $\coprod_ff\circ \epsilon^0_g \neq \epsilon^1_f \circ h$.
Then [1] is a Pullback, and $\coprod_f f: \coprod_f d_0(f) \to \coprod_f d_1(f)$ classifying any arrow of $\mathcal{C''}$.
