To answer the question in the title, I would say that one problem with no-symbols reasonings is that one need to use a lot of pronouns. Problem is, pronouns usually leave too much ambiguity. At some point, mathematical sentences where written without symbols; the introduction of letters to denote mathematical object improved quite a lot the depth and intricacy of reasoning available and communicable.
While it is sometimes possible to write mathematical statements and proofs with few to no mathematical symbols (and I do prefer statement using fewer math symbols), this is not always reasonably possible, and always requires to be very precise (notably with grammar). Your example is telling in this regard: you confound commutation of two representations (which does not make much sense in the context) and commutation of images of two elements under one given representation. This makes your proposition quite problematic.
Even written with great precision, a proof without math symbols can be difficult to decipher, because the reader also needs to be hyper precise in reading it. The slightest slip in the interpretation of which pronoun refers to which object, and the meaning is lost.
That said, the habit of using symbols can sometime impose itself too heavily, and it is often possible to improve a proof by making it lighter, changing a sentence, limiting the introduction of notation. But the question is not math symbols versus flowery language, it is about how to best convey the proof to your intended audience. This is a craft that can be improved without limit.
Also, note that separating the statement from its proof is very important, as they play very different roles. You want the statement to be prominent, because it will usually be referred to several times.
Last, let me give a try at the proof you give in example.
Proof (most natural version to me). Let $U$ be an irreducible representation of an Abelian group $G$ and consider any $p\in G$. Since $G$ is Abelian, for all $g\in G$ we have
$$U(p)U(g) = U(pg) = U(gp) = U(g)U(p)$$
and the operator $U(p)$ commutes with $U$. By Schur's lemma, this implies that $U(p)$ is a scalar operator. Since this is true for all $p$, $U$ is a scalar representation and, being irreducible, it must have dimension $1$.
Proof (attempt without symbols). Given an irreducible representation of an Abelian group, the operator associated with any element of the group must commute with the representation. By Schur's lemma, this implies that this operator is scalar. Since this holds for all elements of the group, the representation is scalar, and being irreducible it must have dimension one.
I do not think much is gained in replacing $G$ by "the group" and $p$ by "element of the group", and in the first proof it was possible to give more details (only suitable for beginners, but possibly crucial for them). Expressing the displayed equations without symbols would be clumsy and certainly unclear. Moreover, this is a very easy proof; try to only state the Five Lemma without maths symbols!