The symmetric algebra is the quotient of the free algebra on the basis by the ideal generated by the relations $e_ie_j-e_je_i$ with $i$, $j\in I$. Pick a linear order $\prec$ on $I$, extend lexicographically to the non-commutative monomials on the generators, find the ambiguities (there are no containment ambiguities, and all overlap ambiguities are of length $3$) You have to check that each of these can be resolved. Do that. Now see what it means in this particular case for a monomial to be in normal form, and observe that the conclusion of the Diamond Lemma is that the set of monomials of the form $e_{i_1}e_{i_2}\cdots e_{i_k}$ with $i_1\preceq i_2\preceq\cdots\preceq i_k$ a non-decreasing sequence of indices with respect to $\prec$. This is the usual basis.
Later. For simplicity, let us suppose that we have finitely many generators $e_1$, $\dots$, $e_n$, and that we order them so that $e_1\prec\cdots\prec e_n$. The relations which define the symmetric algebra are $e_ie_j-e_je_i, \qquad 1\leq j Now, if $1\leq j, the leading term of relation $e_ie_j-e_je_i$ is $e_ie_j$, so the rewriting rules (in Bergman's paper these the pairs $(\sigma,f_\sigma)$...) are $e_ie_j\leadsto e_je_i, \qquad 1\leq j To find the ambiguities, we need to see in what ways the monomials in the set $M=\{e_ie_j:1\leq j interact. There are no containment ambiguities, because they all have length $2$, so we just need to find the overlaps. An overlap of two monomials of length $2$ can only happen in a monomial of length $3$. If $e_ie_je_k$ is such an overlap, we must have $e_ie_j\in M$ and $e_je_k\in M$. We thus conclude that the ambiguitites that we have to resolve are $e_ie_je_k, \qquad 1\leq k