The following seems too easy: maybe I've made a mistake. But here goes.

Write $M^{\mathrm{op}}$ for $M$ with the reverse ordering. Since the negation map $a \mapsto -a$ is an order-reversing involution, it defines an isomorphism $M^\mathrm{op} \to M$. But $M$ is a meet-semilattice with least element, so $M$ is also a join-semilattice with greatest element. Hence $M$ is a lattice (bounded, if that's not already in your definition of "lattice"). Moreover, the isomorphism $a \mapsto -a$ interchanges joins and meets, i.e. the de Morgan laws hold.

Taking $b = -a$ in your condition, we have $a \wedge (-a) = 0$ for all $a$. But by the de Morgan laws, we also have the dual of your condition: $a \vee b = 1$ if and only if $-b \leq a$. Taking $a = -b$ in this dual condition gives $(-b) \vee b = 1$ for all $b$, as required.

I don't know a reference, but that argument (if correct) can't possibly be new.

Edit: Patricia Hersh points out in the comments that I didn't prove it was a Boolean algebra: I only showed that it was a complemented lattice. But maybe this fragment is useful, so I'll leave it here.

Write $M^{\mathrm{op}}$ for $M$ with the reverse ordering. Since the negation map $a \mapsto -a$ is an order-reversing involution, it defines an isomorphism $M^\mathrm{op} \to M$. But $M$ is a meet-semilattice with least element, so $M$ is also a join-semilattice with greatest element. Hence $M$ is a lattice (bounded, if that's not already in your definition of "lattice"). Moreover, the isomorphism $a \mapsto -a$ interchanges joins and meets, i.e. the de Morgan laws hold.

Taking $b = -a$ in your condition, we have $a \wedge (-a) = 0$ for all $a$. But by the de Morgan laws, we also have the dual of your condition: $a \vee b = 1$ if and only if $-b \leq a$. Taking $a = -b$ in this dual condition gives $(-b) \vee b = 1$ for all $b$, as required.