Sorry to disappoint, but if you want to "algebraically" prove the equivalence, you are going to need to use distributivity and DeMorgan's, and a couple more identities.
Since each side of the equivalence has the term $\;(\lnot y \land \lnot z) \lor,\;$
we can first focus on:
$$x \land ((\lnot y \land z) \lor (y \land \lnot z))) \equiv x\land (\lnot (y \land z)))$$
And now you see each side has an equivalent conjunct: $x \land\;$
Indeed, to prove: $$(\lnot y \land \lnot z) \lor (x \land (\lnot y \land z) \lor (y \land \lnot z))) \equiv (\lnot y \land \lnot z) \lor (x \land (\lnot (y \land z)))$$
we first focus on the following to check for equivalence:
$$\lnot (y \land z) \equiv^? (\lnot y \land z) \lor (y \land \lnot z)$$
There is no getting around using the distributive properties, absorption, and DeMorgan's, etc. but this breaks it down so as to make such applications relatively transparent.
$$\lnot (y \land z)\; \equiv^? \;(\lnot y \land z) \lor (y \land \lnot z)$$
$$\equiv\; [(\lnot y \land z) \lor y] \land [(\lnot y \land z )\lor \lnot z]\tag{distributivity}$$
$$\equiv [(\lnot y \lor y) \land (z \lor y)] \land [(\lnot y \lor \lnot z) \land (z \lor \lnot z)]\tag{?}$$
$$ \equiv T \land (z \lor y) \land (\lnot y \lor \lnot z) \land T\tag{?}$$
$$\equiv (z \lor y) \land (\lnot y \lor \lnot z)\tag{?}$$
$$ \vdots$$
$$ \vdots$$
Don't forget that each side of the original equivalence is also also true whenever $\lnot y \land \lnot z \equiv \lnot (y \lor z)$ is true, as it is a disjunct to the expressions on each side, so if it holds, the equivalence holds.
So, e.g., if $x$ is false, then the expressions on the RHS and LHS are equivalent because then, the truth values of each expression is equivalent to the truth value of $\lnot p \land \lnot q$: i.e., each side evaluates to the equivalent truth value.
To prove algebraically, just keep the "disjunct" and $x \land$ "riding along" (operating only as you would above) until you can safely eliminate $x$ and/or make use of $\lnot p \land \lnot q$ to establish the equivalence.
Final Note Using the "$=$"-sign isn't really appropriate in logic. I have used the symbol for logical equivalence: "$\equiv$", which means (equivalently) $\iff$: "if and only if". It can be viewed as a logical connective in its own right. I'll include the truth-table displaying its truth-functional definition:
Best Answer
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
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.