By definition, $\rm b\:$ is a complement of $\rm a\:$ if $\rm\ a\vee b = 1,\ a\wedge b = 0\:$. So a unique complement must be a unique solution to both$\ $ equations (involving both$\ $ operations), not just a single operation - as you consider above. So there is no contradiction.
It seems like you already deduced that
$$x\wedge x = x,\quad
x\wedge y = y \wedge x,\quad
x\wedge(y\wedge z) = (x\wedge y)\wedge z$$
are identities satisfied by the algebra.
(And also that $a\wedge b=a$ iff $a\leq b$.)
Now consider the result
Theorem (O. Frink).
Let $\mathbf A = (A,\cdot,',0)$ be an algebra of type $(2,1,0)$ (that is, $\cdot$ is binary, $'$ is unary, and $0$ is nullary), such that
$(1)\quad xx=x$,
$(2)\quad xy=yx,$
$(3)\quad (xy)z=x(yz),$
$(4)\quad xy=x$ iff $xy'=0$.
Define $\mathbf B = (A,\cdot,+,',0,1)$, where $x+y=(x'y')'$ and $1=0'$.
Then $\mathbf B$ is a Boolean algebra.
The original proof is in
O. Frink, Representations of Boolean algebras, Bulletin Amer. Math. Soc. 47 (1941) 775-776.
An alternative proof (without using duality) can be found in
R. Padmanabhan, A first order proof of a theorem of Frink, Algebra Universalis, 13 (1981) 397-400.
Here, there is an explicit proof of the distributivity.
Now you only have to prove that your algebra satisfies condition $(4)$ of Frink's theorem.
Using your conditions (5) and (6), if $a$ and $b$ are members of the algebra, then
$$ab=a \Leftrightarrow a \leq b \Leftrightarrow a \leq b'' \Leftrightarrow ab'=0,$$
and so indeed, the algebra satisfies all the hypothesis in the theorem.
Best Answer
Every distributive lattice is isomorphic to a lattice of sets, so in particular it is a sublattice of a Boolean algebra. Since Heyting algebras are distributive lattices, the answer to your question is affirmative.