Presentation of the trivial group

Dumb question: Is $\langle e \mid e\rangle$ a presentation of the trivial group?

Yes. What exactly $\langle X|R\rangle$ means? We start with a free group $F(X)$ on $X$ and then we take the quotient $F(X)/N(R)$ where $N(R)$ is the normalizer of some subset $R\subseteq F(X)$.

So what is $\langle e| e\rangle$? While the "$e$" notation may be misleading (a typical symbol for identity) this is the same as $\langle g|g\rangle$, just by simple relabelling. And formally this is the same as $\langle \{g\}|\{g\}\rangle$ which by definition is the same as $F(g)/N(g)$. But since $g$ generates $F(g)$ and $\langle g\rangle\subseteq N(g)$ then $N(g)=F(g)$. And so the quotient $F(g)/N(g)$ is trivial.

Side note: When we write presentations, e.g. $\langle x,y\ |\ xy\rangle$ we often write it in a more readable way, e.g. $\langle x,y\ |\ xy=e\rangle$. And so we introduce the neutral element "$e$" symbol to see it better, but formally we don't actually need this neutral element directly. And so care has to be taken when you use the same symbol for a generator and the neutral element. Ideally you would never do that. Otherwise you are likely to run into problems, for example we can write $\langle g|g\rangle$ as $\langle g|g=e\rangle$ but we cannot write $\langle e|e\rangle$ as $\langle e|e=e\rangle$ because of the ambiguity. But we can write $\langle e|e\rangle$ as $\langle e|e=1\rangle$ if we denote the neutral element by "$1$". However you won't see such presentations often, it doesn't feel natural.