Let $L$ be a finite lattice with least element $\hat{0}$, greatest element $\hat{1}$, and Möbius function $\mu$.

Question 1: What class of lattices the following property characterizes? $$\mu(\hat{0},a)=\mu(\hat{0},\hat{1})\mu(a,\hat{1}), \ \forall a \in L$$ It follows that $\mu(\hat{0},\hat{1}) = \pm1$.

Remark: It is satisfied by any boolean lattice, more generally by the face lattice of any convex polytope (as suggested by John Shareshian), and more generally by any Eulerian lattice.
Proof: An Eulerian lattice is a graded lattice $L$ such that for any $a,b \in L$ with $a \le b$, we have $\mu(a,b)= (-1)^{|b|-|a|} $, with $a \mapsto |a|$ the rank function. The result is immediate. $\square$

Question 2: Is there a non-Eulerian lattice with the above property on the Möbius function?
Yes, see the answer of John Machacek.

As suggested by Sam Hopkins:
Question 3: Is there a non-Eulerian atomistic lattice with the above property on the Möbius function?
No for $|L| \le 13$, as checked by the following Sage program (using these lists of Martin Malandro):

from itertools import product
def relationtest(L,n):
    for l in L:
        P=Poset((range(n),l))
        b = P.bottom()
        t = P.top()
        if all(P.moebius_function(b,x) == P.moebius_function(b,t)*P.moebius_function(x,t) for x in P):
            L=LatticePoset(P)
            if L.is_atomic():
                if not L.is_graded():
                    print(P.cover_relations())
                if L.is_graded():
                    for x, y in product(P, P):
                        if P.compare_elements(x,y)==-1:
                            if not P.moebius_function(x,y)==(-1)^(P.rank(y)-P.rank(x)):
                                print(P.cover_relations())
                                break

Are the small atomistic lattices listed somewhere?