Let $A,B$ be finite SI-algebras and $V(A)$ be congruence-distributive. Then $V(A) = V(B) \iff A \simeq B$
$"\Rightarrow"$ Assume that $V(A) = V(B)$. By a corollary by Jónsson, all the SI-algebras from $V(A)$ are in $HS(A)$. and $V(A) = IP_S(HS(A))$. By the assumption we have $ V(A) = IP_S(HS(A))= IP_S(HS(B))$ $ = V(B)$. Now, a variety is generated by the SI-algebras within it. I am lost how to use it from here to get the desired result.
$"\Leftarrow"$ Assume that $A \simeq B$. Then $\forall C \le A, \exists D \le B$ such that $C \simeq D.$ And, for all $n\in \mathbb{N}$, $A^n \simeq B^n$. Finally $\forall A_H \in H(A), \exists B_H \in H(B)$ such that $A_H \simeq B_H$. Hence, up to isomorphism every algebra in $V(A)$ is similar to some algebra of $V(B)$, and vice versa. Thus, $V(A) = V(B)$. This direction seems like it is relatively trivial when considering that $V(A) = HSP(A)$.
[$\Rightarrow$] $A$ is SI in $V(A)=V(B)$, so $A\in HS(B)$. This means that there is a subalgebra $B'\leq B$ and a surjective homomorphism $\varphi\colon B'\to A$. Now either $B'=B$ and $\varphi$ is an isomorphism of $B$ onto $A$, or else $|A|<|B|$. Now run the argument the other direction: $B$ is SI in $HS(A)$, so either there is an isomorphism of $A$ onto $B$, or else $|B|<|A|$. We can't have both $|A|<|B|$ and $|B|<|A|$, so from one case or the other we get an isomorphism between $A$ and $B$.
[$\Leftarrow$] If $A\cong B$, then $A\in V(B)$ and $B\in V(A)$, since varieties are closed under the formation of isomorphic images. Hence $V(A)\subseteq V(B)$ and $V(B)\subseteq V(A)$, yielding $V(A)=V(B)$.