Yes, at least if your chosen element is the lexicographically first reduced expression for your permutation. This is how people usually prove that any two reduced expressions for a given permutation are connected by a series of long and short braid moves -- by reducing both to the lexicographically first reduced expression for that permutation. This is done by greedily applying $s_j s_i \rightarrow s_i s_j $ for $|j-i|>1$ and $s_{i+1}s_is_{i+1} \rightarrow s_is_{i+1}s_i $ moves until no further such moves are possible.
You can find such an algorithm for the symmetric group e.g. in:
Adriano Garsia, The saga of reduced factorizations of elements of the symmetric group, Publications du Laboratoire de Combinatoire et d' Informatique 29 (2002)
While oddly enough, I don't see this book listed on MathSciNet, I just found a copy available for free by googling: Adriano Garsia saga
This is also discussed (for more general Coxeter groups) in the book:
Anders Björner and Francesco Brenti, Combinatorics of Coxeter groups, Graduate Texts in Mathematics, 231, Springer, New York, 2005, xiv + 363 pp.
At least the connectedness result is there, and I believe it is again by this type of algorithm -- I don't have the book with me right now to check for sure.
Another potentially relevant reference is:
Paul Edelman, Lexicographically first reduced words, Discrete Math, 147 (1995), no. 1-3, 95--106.
Edit: regarding nonreduced words, I should have also said that this greedy algorithm will always produce consecutive letter $s_i s_i$ if your word is not reduced, enabling your other moves $s_i s_i \rightarrow e$, so you will get the desired lexicographically first reduced word. This follows from the fact that the greedy algorithm will transform every reduced subword of consecutive letters to its lexicographially first representative, or else the algorithm will not yet have terminated.