Is monadicity preserved by the underlying functor?

Following up on my comment, I think the direct proof is easiest - it’s just an exercise in translating classical definitions into the enriched setting. Let’s call the underlying monad $T_0$.

For objects, observe that an algebra of $T$ is given by $a: I \to V(TA, A)$ (which is a map $TA \to A$ in the underlying category) so that $\rho^{-1};(\eta_A \otimes a);m = j_A$ (which is exactly $\eta_A;a = id $ in the underlying category) and $\rho^{-1};((a;T) \otimes a);m= \rho^{-1};(\mu_A \otimes a);m$ (which is precisely $T_0(a);a = \mu_A;a$ in the underlying category).

Let $(A,a), (B,b)$ be algebras if T. A map $f:a \to b$ is a point $f:I \to V(A,B)$ so that $T_0(f)b = af$, that is $f;\rho^{-1};(T \otimes b)m = f;\lambda^{-1};(a \otimes id )m$. This is equivalent to saying $f$ is a point of $eq(\rho^{-1};(T \otimes b)m, \lambda^{-1};(a \otimes id )m) = eq(T;V(TA,b), V(a,B))$ which is the hom-object between the algebras.