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
First, I don't know of any name given to this principle in the context of boolean algebra, though if I had to give it a name, it would probably be 'Generalized Reduction' rather than 'Generalized Absorption'
Here are the normal Absorption and Reduction Principles:
Absorption
$P \land (P \lor Q) \Leftrightarrow P$
$P \lor (P \land Q) \Leftrightarrow P$
Reduction
$P \land (\neg P \lor Q) \Leftrightarrow P \land Q$
$P \lor (\neg P \land Q) \Leftrightarrow P \lor Q$
So what you do is more like Reduction than Absorption: rather than that the whole second term is being 'absorbed' (and thus completely eliminated) by the single $P$, the second term is being 'reduced' (and thus simplified) in the context of the single term $P$
Second, and I think a little more interestingly, there is a visual logic system, called Existential Graphs, whose Iteration/Deiteration rule(s) very closely relate to your principles.
Quick explanation of Existential Graphs:
Statements in this system are put onto a Sheet of Assertion. It does not matter where you put the statements ... anything put on the Sheet of Assertion is being asserted as true. Thus, for example:
asserts $P$ as well as $Q$.
There is no explicit conjunction symbol in EG, but you can treat the fact that two statements are being juxtaposed in the same area as a conjunction. Thus, the above graph can be interpreted as having asserted two separate statements $P$ and $Q$, but also as having asserted a single statement $P \land Q$, or (since location does not matter) as the single statement $Q \land P$. This means that there is no one-to-one mapping between graphs in EG and algebraic expressions in normal logic systems, but it is actually quite a powerful cognitive feature: no conjunction symbol is needed, no order is imposed on the statements, and the Commutation and Association principles come for free.
Now, to negate a statement, we do this:
The enclosed figure is called a 'cut'. Putting a cut around some expression negates that expression. So, for example:
means the negation of the juxtaposition (conjunctin) of $P$ and $Q$, i.e. in normal notation this would be either $\neg (P \land Q)$ or $\neg (Q \land P)$
Now, since $\{ \land, \neg \}$ is an expressively complete set, this is actually all we need: the letters and cuts are all we need to express any logic statement. In fact, the cut can be seen as a generalized NAND: it negates everything that is conjuncted inside of it. But in practice, it typically works better to interpret the cut as a negation.
Here is a disjunction $P \lor Q$:
And here is an implication $P \rightarrow Q$:
This is a very common structure: when you have one cut inside another, then the stuff that is between the cuts (i.e. inside the outside cut, but outside the inside cut) is the antecedent, and the stuff that is inside the inside cut is the consequent. Thus, for example, the graph for the disjunction above can also be read as $\neg P \rightarrow Q$ or as $\neg Q \rightarrow P$ ... showing once again the power of these graphs: many equivalence principles become immediately obvious, and many principles intuitively generalize.
... which brings us to your rules:
$$P + (\ldots P \ldots) = P + (\ldots 0 \ldots)$$
and
$$P \cdot (\ldots P \ldots) = P \cdot (\ldots 1 \ldots)$$
Now, first it should be pointed out the in Existential Graphs you do not have an explicit $0$ or $1$ (or $\bot$ and $\top$). However, you can treat any empty bit of space as a $1$ (or $\top$): since all the bits on a graph are implicitly being conjuncted, then the any graph can be seen as that same graph 'conjuncted' with any empty bit of space next to it. But what logic statement has the feature that, when conjuncted with any other statements, results in that other statement? The $1$ and $\top$ of course!
Note this this also immediately means that:
is the $0$ (or $\bot$). This is called the 'empty cut'.
OK, now it gets interesting. Existential Graphs has 4 inference rules defined:
and:
Thus, for example, we can go from:
to:
Now, remembering that we can read the original graph as the implication $P \rightarrow (Q \land R)$, and the resulting graph as $P \rightarrow Q$, we recognize this particular inference as the valid 'Weakening the Consequent' inference principle ... but again, the compact nature of Existential Graphs, which allows for many alternative readings, means that the Erasure rule is really much more general than that; it is a kind of 'Generalized Weakening the Consequent'.
OK, and now we come to the interesting rule that I think bears a lot of resemblance to your rules:
Let's do Modus Ponens as an example. Start with $P$ and $P \rightarrow Q$:
Now, the $P$ inside the cut is at a 'nested' level with regard to the outside $P$ , so we can 'deiterate' it:
Now we can use Double Cut:
And, if you don't like the fact that the $P$ is still there, we can simply remove it by Erasure (level $0$ is even):
OK, so now let's finally make the connection to your rules. First, consider:
$$P \cdot (\ldots P \ldots) = P \cdot (\ldots 1 \ldots)$$
Well, the graph for the left side would be:
(the dots indicate that the right $P$ can occur at any nested level, even if it is many levels deeper than the outside $P$)
while the right hand side would be:
(remember that a $1$ corresponds to a bit of empty space in Existential Graphs)
OK ... so that is exactly the Iteration/Deiteration rule which, since it goes both ways, is an equivalence principle (just like Double Cut, but unlike Erasure and Insertion, which are one-way inference rules)
OK, now for the other one:
$$P + (\ldots P \ldots) = P + (\ldots 0 \ldots)$$
This one is a little more tricky, but notice that the left hand side is the following graph:
So, if we now do a Double Cut on the inside $P$:
Then we can Deiterate:
.. and remembering that the $0$ corresponds to the Empty Cut, we're there. And, of course, using Double Cut and Iteration we can go back, demonstrating the equivalence.
... So .... I would say that there is a definite connection between your principles and EG's Iteration/Deiteration (equivalence) rule!