$\newcommand\de\delta\newcommand\ep\varepsilon$The conjecture is true.

Indeed, let $(z_n)$ be a Cauchy-convergent sequence in $(Z,d)$. Then $$d_{W_p}(\de_{z_m},\de_{z_n})=d(z_m,z_n)\to0$$ (as $m,n\to\infty$). So, the sequence $(\de_{z_n})$ is Cauchy-convergent and hence converges in $(W_p,d_{W_p})$ to some probability measure $\mu$ over $Z$.

That is, $\int_Z \mu(dz)\,d(z,z_n)^p\to0$. So, by Markov's inequality, $1-\mu(B_\ep(z_n))=\int_Z \mu(dz)\,1(d(z,z_n)\ge\ep)\to0$ for each real $\ep>0$, where $B_\ep(z)$ denotes the open ball of radius $\ep$ centered at $z$. It follows that, if $u$ is any point in the support set $S_\mu$ of the probability measure $\mu$, then $z_n\to u$. So, any Cauchy-convergent sequence $(z_n)$ in $(Z,d)$ is convergent. $\quad\Box$

$\newcommand\de\delta\newcommand\ep\varepsilon$The conjecture is true.

Indeed, let $(z_n)$ be a Cauchy-convergent sequence in $(Z,d)$. Then $$d_{W_p}(\de_{z_m},\de_{z_n})=d(z_m,z_n)\to0$$ (as $m,n\to\infty$). So, the sequence $(\de_{z_n})$ is Cauchy-convergent and hence converges in $(W_p,d_{W_p})$ to some probability measure $\mu$ over $Z$.

That is, $\int_Z \mu(dz)\,d(z,z_n)^p\to0$. So, by Markov's inequality, $1-\mu(B_\ep(z_n))=\int_Z \mu(dz)\,1(d(z,z_n)\ge\ep)\to0$ for each real $\ep>0$, where $B_\ep(z)$ denotes the open ball of radius $\ep$ centered at $z$. It follows that, if $u$ is any point in the support set $S_\mu$ of the probability measure $\mu$, then $z_n\to u$. So, any Cauchy-convergent sequence $(z_n)$ in $(Z,d)$ is convergent. $\quad\Box$

This result holds for $p=\infty$ as well. Indeed, in this case the $\mu$-essential supremum of the function $d(\cdot,z_n)$ goes to $0$. So, again, if $u$ is any point in the support set $S_\mu$ of the probability measure $\mu$, then $z_n\to u$.