Yes; regularity and $\in$-induction are equivalent, in fact by first-order logic alone! To see this, it’s easiest to speak explicitly in terms of classes.
In these terms, $\in$-induction says: “For every class $C$, if $C$ is $\in$-hereditary, then $C$ contains all sets.” Regularity says: “For every class $D$, if $D$ is inhabited, then $D$ has an $\in$-minimal element.”
But “$\in$-hereditary” is dual to “has an $\in$-minimal element”, in the sense that a class $C$ is hereditary precisely if its complement $\bar{C}$ does not have an $\in$-minimal element, and “contains all sets” is dual to “is inhabited” in the same sense.
So we have a chain of equivalent statements, using purely first-order logic:
- If $C$ is $\in$-hereditary, then $C$ contains all sets
- (contrapositive) If $C$ does not contain all sets, then $C$ is not $\in$-hereditary.
- (duality as noted above) If $\bar{C}$ is inhabited, then $\bar{C}$ has an $\in$-minimal element.
So this shows: the instance of $\in$-induction for any class $C$ is equivalent to the regularity for its complement $\bar{C}$, and vice versa.
Going back to the first-order axiom-scheme versions, we have that an arbitrary instance of $\in$-induction, for the formula $\varphi(x,y_1,\ldots,y_n)$, is equivalent to the instance of regularity for the complementary formula $\lnot \varphi(x,y_1,\ldots,y_n)$; and similarly, every instance of regularity for a formula $\varphi$ is equivalent to the instance of $\in$-induction for $\lnot \varphi$.