Python 3, 109 bytes

from datetime import*
i=input()
z=date(*map(int,i.split())).toordinal()-727806
print([i,'1993 09 %d'%z][z>9])

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-59 bytes thanks to notjagan
-3 bytes thanks to Mr. Xcoder
-2 bytes thanks to officialaimm
-12 bytes thanks to Jonathan Allan

• \$\begingroup\$ -54 bytes. \$\endgroup\$

  notjagan

  – notjagan

  2017年08月06日 04:49:05 +00:00

  Commented Aug 6, 2017 at 4:49
• 1
  \$\begingroup\$ Or better yet, -59 bytes. \$\endgroup\$

  notjagan

  – notjagan

  2017年08月06日 05:01:29 +00:00

  Commented Aug 6, 2017 at 5:01
• 1
  \$\begingroup\$ 123 bytes (-62) \$\endgroup\$

  Mr. Xcoder

  – Mr. Xcoder

  2017年08月06日 07:27:21 +00:00

  Commented Aug 6, 2017 at 7:27
• 1
  \$\begingroup\$ -8644+1 can be -8643.. \$\endgroup\$

  0xffcourse

  – 0xffcourse

  2017年08月06日 07:45:25 +00:00

  Commented Aug 6, 2017 at 7:45
• 1
  \$\begingroup\$ @Mr.Xcoder Needs to be z>9 otherwise you lose the leading zero on the day. \$\endgroup\$

  Neil

  – Neil

  2017年08月06日 10:37:19 +00:00

  Commented Aug 6, 2017 at 10:37

JavaScript (ES6), 48 bytes

f=
s=>(d=new Date(s)/864e5-8643|0)>9?'1993-09-'+d:s

<input size=10 oninput=o.textContent=/\d{4}(-\d\d){2}/.test(this.value)?f(this.value):``><pre id=o>

Based on @Mr.Xcoder's algorithm.

Mathematica, 55 bytes

If[(s=#&@@{1993,9}~DateDifference~#)>0,{1993,9,s+1},#]&

I/O

{2017, 8, 6} ->{1993, 9, 8741}
{1986, 1, 28}->{1986, 1, 28}

-6 bytes thanx to user202729

• \$\begingroup\$ Would you consider shift the time mark {1993,9,1} back by a day, so as to remove the +1, saving 2 bytes? \$\endgroup\$

  user202729

  – user202729

  2017年08月06日 15:07:01 +00:00

  Commented Aug 6, 2017 at 15:07
• \$\begingroup\$ Thanks. I should try to be more polite next time. And I don't even know {1993,9,0} is allowed. \$\endgroup\$

  user202729

  – user202729

  2017年08月06日 15:22:43 +00:00

  Commented Aug 6, 2017 at 15:22

Perl 5, 102 + 16 (-MTime::Local -F-) = 118 bytes

,ドル='-';say @F=($t=timelocal(0,0,0,$F[2],$F[1]-1,$F[0]-1900)-749433599)>0?(1993,'09',31+int$t/86400):@F

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Takes the date as "YYYY-MM-DD"

I think I did the count right on the command line options. I'm sure someone will correct me if I didn't.

C# (.NET Core), 107 bytes

s=>{var d=(System.DateTime.Parse(s)-new System.DateTime(1993,8,31)).TotalDays;return d<1?s:"9/"+d+"/1993";}

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Takes dates as M/D/YYYY (numbers below 10 written with only 1 digit). Written from my mobile phone using the API by heart.

Gaia, 78 bytes

HZ¤∨4Ė
:'//d¦[1993931];>\{\‡:(...1993>↑¦365+¦¤ṇ↑∂K∂k,=;((<¤)-243]_ḥΣ"1993/09/"¤+}?

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Explanation

First, we have a helper function that determines if a year is a leap year.

H 100
 Z Divmod year by 100, pushing the first 2 digits, then the second 2 digits
 ¤ Swap
 ∨ Logical OR, gives the left-most non-zero number
 4Ė Check for divisibility by 4

The main function does the rest of the work:

: Push two copies of the input.
'// Split the top on on slashes.
d¦ Parse each section as a number.
[1993931] Push the list [1993 9 31].
;> Copy the date and check if its less than that.
\ If it is, delete the list and leave the input string on top.
{ Else:
 :( Copy the date and get the year.
 ...1993> Get the range from 1993 to year-1.
 ↑¦365+¦ Map each to 365+(whether it's a leap year).
 ¤ Swap, bring the date back to the top.
 ṇ↑ Pull out the year and check if it's a leap year.
 ∂K∂k, Push the pair of lists [[days in months in a leap year] [days in months]]
 = Index the result of checking if the year is a leap year into the pair.
 ;((< Get the first (month number - 1) elements.
 ¤ Swap, bring date back to the top.
 ) Get the day.
 -243 Push -243 (243 is the number of days between Jan 1 1993 and Sept 1 1993).
 ] Wrap everything in a list.
 _ Flatten the list.
 ḥ Remove the first element (the input string).
 Σ Sum it.
 "1993/09/"¤+ Append the resulting number to "1993/09/".
}? (end if)
 Implicitly display whatever is on top of the stack.

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