Two lattice identities $$a=(b \land a) \lor a$$
$$(((a \land b) \land c) \lor d) \lor e = (((b \land c) \land a) \lor e) \lor ((f \lor d) \land d)$$
To be shown equivalent to Absorption, Idempotency, Commutativity, and Associativity and of both operations.
I'm interested in a proof that uses nothing else (And there is a lot to use).
Question: Can I just use self duality of lattices?  
 A: Although they're not exactly the same identities (the absorption one coincides), you might be also interested in the paper Two identities for lattices by R. Padmanabhan (Proc. AMS, vol 20 (1969), pages 404-412). Here, the other identity (out of the same absorption law) is
$$((x \wedge f) \wedge z) \vee (((a \vee u) \wedge u) \vee v) \vee w 
= ((g \wedge z) \wedge x) \vee ((v \vee w) \vee u).$$ 
Also of interest (and we shall use it below), is the following result, also from Padmanabhan, from the paper On axioms for semi-lattices (Canad. Math. Bull., vol 9 (1966) pages 357-358):

Lemma.
  Let $\mathbf S = \langle S, \cdot \rangle$ be an algebra (where $\cdot$ is a binary operation) that satisfies the axioms
  $$xx=x, \quad (xy)z = (yz)x.$$
  Then $\mathbf S$ is a semi-lattice.
Proof. First, $\mathbf S$ is commutative:
  $$xy = (xy)(xy) = (y(xy))x = ((xy)x)y = ((yx)x)y = ((xx)y)y = (xy)y = (yy)x = yx.$$
  Now, 
  $$(xy)z = (yz)x = x(yz).$$

Let us now tackle the problem.
Below follows a proof of the traditional axioms for lattices (for each operation, is a semi-lattice and the absorption laws).
The laws of idempotency are redundant, but very handy.
There are a lot of other identities which are deduced in the process, and used to prove the main ones.
I will highlight the ones we're interested in.
These (commutativity, associativity, idempotency and absorption) shall be used mostly without explicit reference to them.
So we start with

\begin{equation}\label{eq:1}
(x \wedge y) \vee y = y,                                                \tag{1}
\end{equation}

and 
\begin{equation}\label{eq:2}
(((x \wedge y) \wedge z) \vee t) \vee u 
= (((y \wedge z) \wedge x) \vee u) \vee ((v \vee t) \wedge t).          \tag{2}
\end{equation}
Equations \eqref{eq:1} and \eqref{eq:2} are the ones in your post.
In \eqref{eq:2}, taking $u=x$ and using \eqref{eq:1}, we get
\begin{equation}\label{eq:3}
(((x \wedge y) \wedge z) \vee t) \vee x = x \vee ((v \vee t) \wedge t), \tag{3}
\end{equation}
and taking $t=z$ (again, using \eqref{eq:1}, but I won't make an explicit reference to it most of the times),
\begin{equation}\label{eq:4}
z \vee x = x \vee ((v \vee z) \wedge z).                           \tag{4}
\end{equation}
In \eqref{eq:3}, taking $v = x \wedge t$,
\begin{equation}
(((x \wedge y) \wedge z) \vee t) \vee x 
= x \vee (((x \wedge t) \vee t) \wedge t) = x \vee (t \wedge t),   \tag{5}
\end{equation}
and taking $z=t$,
\begin{equation}\label{eq:6}
t \vee x = x \vee (t \wedge t).                                    \tag{6}
\end{equation}
By \eqref{eq:6} with $t=x$ and $x=y \wedge (x \wedge x)$,
\begin{equation}\label{eq:7}
x \vee (y \wedge (x \wedge x)) 
= (y \wedge (x \wedge x)) \vee (x \wedge x) = x \wedge x.           \tag{7}
\end{equation}
Now, taking $y= x \wedge x$,
$$x \wedge x = x \vee ((x \wedge x) \wedge (x \wedge x)),$$
by \eqref{eq:7}.
Applying \eqref{eq:6}, we get
$$x \vee ((x \wedge x) \wedge (x \wedge x)) = (x \wedge x) \vee x = x,$$
and thus,

\begin{equation}\label{eq:8}
x \wedge x = x.                                                    \tag{8}
\end{equation}

By \eqref{eq:1}, $(x \wedge x) \vee x = x$, and so by \eqref{eq:8},

\begin{equation}\label{eq:9}
x \vee x = x.                                                      \tag{9}
\end{equation}

In \eqref{eq:4}, taking $z=x$ we get
\begin{equation}\label{eq:10}
x = x \vee x = x \vee ((v \vee x) \wedge x);                      \tag{10}
\end{equation}
also from \eqref{eq:4},
$$x \vee ((v \vee t)\wedge t) 
= ((v \vee t) \wedge t) \vee ((y \vee x) \wedge x),$$
and taking $t=x$ and $v=y$,
\begin{equation}\label{eq:11}
x \vee ((y \vee x) \wedge x) 
= ((y \vee x) \wedge x) \vee ((y \vee x) \wedge x) 
= ((y \vee x) \wedge x).                                          \tag{11}
\end{equation}
Hence, from \eqref{eq:10} and \eqref{eq:11},

\begin{equation}\label{eq:12}
x = (y \vee x) \wedge x.                                          \tag{12}
\end{equation}

From \eqref{eq:4} and \eqref{eq:12},

\begin{equation}\label{eq:13}
x \vee z = z \vee x.                                              \tag{13}
\end{equation}

In \eqref{eq:2}, taking $t=v$,
\begin{equation}\label{eq:14}
(((x \wedge y) \wedge z) \vee v) \vee u
=(((y \wedge z) \wedge x) \vee u) \vee v,                         \tag{14}
\end{equation}
and making $z=y$,
\begin{equation}
(((x \wedge y) \wedge y) \vee v) \vee u
=((y \wedge x) \vee u) \vee v.                                    
\end{equation}
Now, with $y=x$,
$$(x \vee v) \vee u = (x \vee u) \vee v,$$
and, by the Lemma (and \eqref{eq:13}),

\begin{equation}\label{eq:15}
(x \vee y) \vee z = x \vee (y \vee z).                             \tag{15}
\end{equation}

From here, it follows that 
\begin{equation}\label{eq:16}
x \vee (x \vee y) = (x \vee x) \vee y = x \vee y.                  \tag{16}
\end{equation}
In \eqref{eq:14} with $v=u$ we get
$$(((x \wedge y) \wedge z) \vee u) \vee u
= (((y \wedge z) \wedge x) \vee u) \vee u,$$
and, by \eqref{eq:16},
\begin{equation}\label{eq:17}
((x \wedge y) \wedge z) \vee u = ((y \wedge z) \wedge x) \vee u.   \tag{17}
\end{equation}
In \eqref{eq:17}, taking $u= (x \wedge y) \wedge z$, we get
$$(x \wedge y) \wedge z = 
((y \wedge z) \wedge x) \vee ((x \wedge y) \wedge z);$$
in \eqref{eq:17}, taking $u = (y \wedge z) \wedge x$, we get
$$((x \wedge y) \wedge z) \vee ((y \wedge z) \wedge x)
= (y \wedge z) \wedge x,$$
whence
$$(x \wedge y) \wedge z = (y \wedge z) \wedge x.$$
By Lemma,

\begin{equation}
x \wedge y = y \wedge x,
\end{equation}

and

\begin{equation}
x \wedge (y \wedge z) = (x \wedge y) \wedge z.
\end{equation}

