Mirrored Digital Clock

Python 2, 187 180 178 177 bytes

R=range(11)
for t in['0000111122201250125012'[j::11]+':'+'0001112255501501501015'[i::11]for i in R for j in R]:print t+' - '+''.join(map(dict(zip('0125:','0152:')).get,t))[::-1]

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Thanks for +1 Kevin Cruijssen.

APL (Dyalog Unicode), 84 bytes^SBCS

Complete program outputting to STDOUT. Requires ⎕IO (Index Origin) to be 0 which is default on many systems.

{0::⋄∧/23 59≥⍎¨(':'≠t)⊆t←⌽'015xx2xx8x:'[⎕D⍳i←∊⍺':'⍵]:⎕←1↓⍕i'-'t}⌿1↓¨⍕¨100+0 60⊤⍳1440

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⍳1440 that many ɩntegers

0 60⊤ convert to mixed-base ∞,60

100+ add 100 (this pads the needed 0s)

⍕¨ format (stringify) each

1↓¨ drop the first character from each (this removes the leading 1s)

{…}⌿ apply the following anonymous function column-wise (⍺ is top hour, ⍵ is minute)

 0:: if any error happens, return nothing

 ⋄ try:

  '015xx2xx8x:'[…] index this string with:

   ∊⍺':'⍵ the ϵnlisted (flattened) list of hour, colon, minute

   i← stored in i (for input)

   ⎕D⍳ ɩndices of each character in the list of Digits

  ⌽ reverse that

  t← store as t (for time)

  (…)⊆ group runs where:

   ':'≠t colon differs from t

⍎¨ execute (evaluate) each

 23 59≥ Boolean for each whether they are less than or equal to 23 and 59 respectively

 ∧/ are both true?

 : if so, then:

  ⍕i'-'t the formatted (space-separated) list of input, dash, time

  1↓ drop the first (space)

  ⎕← output to STDOUT

Python 2, 279 277 255 bytes

for h in range(1440):
 q=[[[0,(a+"52")[(a=="2")+(a=="5")*2]][a in"01825"]for a in c]for c in[("%02d"%e)[::-1]for e in[h%60,h/60]]]
 if all(q[0]+q[1]):
	z=[int(''.join(j))for j in q]
	if(z[1]<60)*(z[0]<24):print"%02d:%02d - %02d:%02d"%(h/60,h%60,z[0],z[1])

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Credits

• 279 bytes reduced to 256 by dylnan.

• 256 bytes reduced to 255 by FlipTrack.