It is not true.
Consider the lattices $\mathbf{M}_3,\mathbf{M}_4, \dots$ only with the language $\{\vee,\wedge\}$ instead of $\{\vee,\wedge,\neg\}$. First of all, one can show that $\mathbf{M}_n$ is simple for $n\geq 3$. Second of all, since the class of all lattices is a congruence distibutive variety, we can use Jonsson's lemma to say that the subdirectly irreducibles in $\mathsf{V}(\mathbf{M}_n)$ belong to $\mathsf{HS}(\mathbf{M}_n)$. Since the only sublattices of $\mathbf{M}_n$ are the lattices $\mathbf{M}_k$ for $k=0,1,2,\dots,n$, and each of those (except $\mathbf{M}_1$ and $\mathbf{M}_2$) are simple, we get that the subdirectly irreducibles in $\mathsf{V}(\mathbf{M}_n)$ are exactly $\mathbf{M}_0,\mathbf{M}_3,\dots,\mathbf{M}_n$.
So $\mathsf{V}(\mathbf{M}_m)$ is strictly contained in $\mathsf{V}(\mathbf{M}_n)$ whenever $3\leq m<n$. So, by Birkhoff's theorem, there must be some universally quantified equation $\forall v, \phi(v)=\chi(v)$ satisfied by $\mathbf{M}_m$ that is not satisfied by $\mathbf{M}_n$. So in $\mathbf{M}_n$ $\exists v,\phi(v)\neq\chi(v)$ (if $\phi(v)<\chi(v)$, then switch $\phi$ and $\chi$). Then $\phi\models_{\mathbf{M}_m}\chi$, but $\phi\not\models_{\mathbf{M}_n}\chi$. Since $\phi,\chi$ are formulas in $\{\vee,\wedge\}$, they are also formulas in $\{\vee,\wedge,\neg\}$.
As a specific example, $\mathbf{M}_3$ satisfies the equation
$
x_1\vee((x_2\vee(x_3\wedge x_4))\wedge(x_3\vee(x_1\wedge x_4)))\approx x_1\vee(((x_1\vee x_2)\vee(x_3\wedge x_4))\wedge(x_3\wedge(x_2\vee x_4)))
$
but $\mathbf{M}_4$ does not. Let
$\chi(x_1,x_2,x_3,x_4)=x_1\vee((x_2\vee(x_3\wedge x_4))\wedge(x_3\vee(x_1\wedge x_4)))$
and
$\phi(x_1,x_2,x_3,x_4)=x_1\vee(((x_1\vee x_2)\vee(x_3\wedge x_4))\wedge(x_3\wedge(x_2\vee x_4)))$
Now $\phi\models_{\mathbf{}_3}\chi$ since $\phi(v)$ always returns the same element as $\chi(v)$, but $\phi\not\models_{\mathbf{M}_4}\chi$ since $\chi(1,2,3,4)=1$ and $\phi(1,2,3,4)=\top$.
The diagrams of $L_1$ and $L_2$ are misleading representations of lattices since they are not Hasse diagrams.
This is because in $L_1$ you have one line to much: the one from $c$ to $b$.
It is redundant because $c < f < b$, whence $c < b$ follows from transitivity.
Erase that line and it becomes obvious that the lattice is distributive (up to isomorphism, it is the lattice $\mathbf 3 \times \mathbf 2$).
Likewise in $L_2$ you have three lines to much: between $c$ and $d$; between $c$ and $e$; and between $b$ and $a$.
Again, erase those and everything becomes clear.
In the sub-lattice definition, the meaning of "same meet and join operations" is that if $a$ and $b$ are in the subset which we are testing if it is a sub-lattice, then for that set to be a sub-lattice, $a \wedge b$ and $a \vee b$ must belong to that set, where these operations are calculated in the main lattice.
So you can have a subset which is a lattice on its own, but not a sub-lattice (for example, in the lattice $L_2$ above, erase $b$, and the resulting subset is a lattice, but it is not a sub-lattice of $L_2$ because $d \wedge e = b$, which doesn't belong to the resulting subset; it is still a lattice where $d \wedge e = c$).
Best Answer
Ok, I think I finally got it right (axioms numbering from here). Commutativity allows us to write it as $$(x \vee y) \vee z = \neg (\neg z \oplus y \oplus \neg(\neg x \oplus y)) \oplus \neg (\neg x \oplus y) \oplus y. $$ From 2., this equals $$= \neg (\neg \neg (\neg z \oplus y)\oplus \neg (\neg x \oplus y))\oplus \neg (\neg x \oplus y) \oplus y. $$ Apply 4. to get $$=\neg (\neg x \oplus y \oplus \neg (\neg z \oplus y))\oplus \neg (\neg z\oplus y) \oplus y. =x \vee (y \vee z)$$