How exactly is an enigma simulation software supposed to encrypt alphabets?
I guess you already figured it out but my guess is that you forgot to move the first rotor 1 step before starting the encryption. That also impacts what letter it arrives to on next rotor and so on.

The complete path is

KBD-A&gt;SB&gt;A - A&gt;ETW&gt;A - B&gt;(1)R3&gt;K - J&gt;(2)R2&gt;B - B&gt;(3)R1&gt;D - D&gt;UKW&gt;H 
  - H&gt;(3)R1&gt;D - D&gt;(2)R2&gt;C - D&gt;(1)R3&gt;G - F&gt;ETW&gt;F - F&gt;SB&gt;F-light


Keyboard A arrives to plugboard (Steckerbrett) as A, leaves the plugboard as A, arrives to Entry wheel as A, leaves as A, arrives to the first rotor at level B, gets rewired to K and so on.
Typing 3 more "A" gives

 A&gt;SB&gt;A - A&gt;ETW&gt;A - C&gt;(1)R3&gt;M - K&gt;(2)R2&gt;L - L&gt;(3)R1&gt;V - V&gt;UKW&gt;W
       - W&gt;(3)R1&gt;R - R&gt;(2)R2&gt;G - I&gt;(1)R3&gt;V - T&gt;ETW&gt;T - T&gt;SB&gt;T  :T

 A&gt;SB&gt;A - A&gt;ETW&gt;A - D&gt;(1)R3&gt;F - C&gt;(2)R2&gt;D - D&gt;(3)R1&gt;H - H&gt;UKW&gt;D
    - D&gt;(3)R1&gt;B - B&gt;(2)R2&gt;J - M&gt;(1)R3&gt;C - Z&gt;ETW&gt;Z - Z&gt;SB&gt;Z  :Z

 A&gt;SB&gt;A - A&gt;ETW&gt;A - E&gt;(1)R3&gt;L - H&gt;(2)R2&gt;U - U&gt;(3)R1&gt;K - K&gt;UKW&gt;N
    - N&gt;(3)R1&gt;N - N&gt;(2)R2&gt;T - X&gt;(1)R3&gt;Q - M&gt;ETW&gt;M - M&gt;SB&gt;M  :M