Yes, you've got it right. Given an unbounded self-adjoint operator $A$ with domain $D(A) \subset H$, using Zorn's lemma you can produce an everywhere defined operator $A'$ on $H$ which extends $A$. (In fact you can produce many such operators; the extension is highly non-unique.) By Hellinger–Toeplitz, $A'$ cannot be symmetric. And you already knew, from the closed graph theorem, that $A'$ cannot be closed. So it's definitely not self-adjoint.

Since you used Zorn's lemma in an essential way, you won't get a "concrete" description of such an $A'$. There's a strong sense in which this is true. A common working definition of "concrete" is "something whose existence you can prove using only the axiom of dependent choice (DC)". There's a famous theorem of Solovay (extended by Shelah) that it's consistent with DC that every set of reals has the property of Baire (BP); i.e. there are models of set theory in which DC and BP both hold (but full AC necessarily fails). But from BP you can prove that every everywhere defined operator on any Banach space is bounded. So in such models, $A$ won't have any extension to all of $H$. Put another way, you can't even prove $A'$ exists without using AC in an essential way, so you certainly can't construct it concretely.

You can read more about these ideas in Schechter's Handbook of Analysis and its Foundations.

Yes, you've got it right. Given an unbounded self-adjoint operator $A$ with domain $D(A) \subset H$, using Zorn's lemma you can produce an everywhere defined operator $A'$ on $H$ which extends $A$. (In fact you can produce many such operators; the extension is highly non-unique.) By Hellinger–Toeplitz, $A'$ cannot be symmetric. So it definitely isn't self-adjoint. Another way to see this that $A'$ cannot be self-adjoint is to note that, by the closed graph theorem, $A'$ cannot be closed.

Since you used Zorn's lemma in an essential way, you won't get a "concrete" description of such an $A'$. There's a strong sense in which this is true. A common working definition of "concrete" is "something whose existence you can prove using only the axiom of dependent choice (DC)". There's a famous theorem of Solovay (extended by Shelah) that it's consistent with DC that every set of reals has the property of Baire (BP); i.e. there are models of set theory in which DC and BP both hold (but full AC necessarily fails). But from BP you can prove that every everywhere defined operator on any Banach space is bounded. So in such models, $A$ won't have any extension to all of $H$. Put another way, you can't even prove $A'$ exists without using AC in an essential way, so you certainly can't construct it concretely.

You can read more about these ideas in Schechter's Handbook of Analysis and its Foundations.