Adding a wait-for-element while performing a SplashRequest in python Scrapy

Yes, you can write a Lua script to do that. Something like that:

function main(splash)
  splash:set_user_agent(splash.args.ua)
  assert(splash:go(splash.args.url))

  -- requires Splash 2.3  
  while not splash:select('.my-element') do
    splash:wait(0.1)
  end
  return {html=splash:html()}
end

Before Splash 2.3 you can use splash:evaljs('!document.querySelector(".my-element")') instead of not splash:select('.my-element').

Save this script to a variable (lua_script = """ ... """). Then you can send a request like this:

yield SplashRequest(
    url, 
    self.parse, 
    endpoint='execute',
    args={
        'lua_source': lua_script,
        'ua': "Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/51.0.2704.106 Safari/537.36"
    }
}

See scripting tutorial and reference for more details on how to write Splash Lua scripts.

I have a similar requirement, with timeouts. My solution is a slight modification of above:

function wait_css(splash, css, maxwait)
    if maxwait == nil then
        maxwait = 10     --default maxwait if not given
    end

    local i=0
    while not splash:select(css) do
       if i==maxwait then
           break     --times out at maxwait secs
       end
       i=i+1
       splash:wait(1)      --each loop has duration 1sec
    end
end