Munkres defines a word as finite sequence $s = (x_1,\dots,x_n)$ of elements $x_i \in G_{\alpha_i}$ and says that such a sequence represents $x \in G$ if $x = \prod_{i=1}^n x_i = x_1\dots x_n$. As an adhoc notation let us write $\prod s = \prod_{i=1}^n x_i$. Then he defines two reduction operations shortening sequences $s$ of length $n$ to sequences $s'$ of length $n-1$ such that $\prod s = \prod s'$. One of these operations is the omission of $x_i$ if $x_i = 1$.
At the beginning he does not say anything about $n$ and thus one might think that we only deal with sequences of length $n \ge 1$. However, after he defined the reduction operations he explicitly makes the convention that the sequence $\emptyset$ of length $0$ (i.e. the empty word) is regarded as a reduced word which represents $1$. This is common practice; for example in elementary calculus one frequently defines the empty sum as $0$ and the empty product as $1$ (i.e. as the neutral elements with respect to additon and multiplication).
Let us come to your question. The empty word always represents $1$. But there may be other reduced words $s$ representing $1$ and then the representation of $1$ as a reduced word is not unique. This is what Munkres means. But admittedly the formulation "the representation of $1$ by the empty word is unique" may be misleading. He should better have said "the representation of $1$ by a reduced word is unique, and this is the representation by the empty word".
Here is an example. Let $G = \mathbb Z \oplus \mathbb Z$. Let $G_1$ be the subgroup of all pairs $(a,0)$ and $G_2$ the subgroup of all pairs $(0,b)$. Their intersection only contains the additive identity element $(0,0)$. Clearly each element $(a,b)$ of $G$ is represented by the word $((a,0),(0,b))$. But all words $((a,0),(0,b),(-a,0),(0,-b))$ with $a,b \ne 0$ are reduced and represent $(0,0)$.
