Keep the entries before column j, discard any old entry at j, and retain those after it. Insert the replacement only when it is nonzero.
def set(i: Int, j: Int, x: Double): Unit = {
  require(0 <= i && i < N && 0 <= j && j < N)
  val before = rows(i).takeWhile(e => e._1 < j)
  val after = rows(i).dropWhile(e => e._1 <= j)
  val middle: Row = if (x == 0.0) Nil else List((j, x))
  rows(i) = before ::: middle ::: after
}
All indices in before are below j and all in after are above it, so the representation invariant is preserved. Other rows and all other column values remain unchanged:
(1)
Assigning zero removes storage for that entry. The time complexity is linear in the stored row length.

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