Let's do some location arithmetic!

JavaScript (ES6), 136 134 133 bytes

Saved 1 byte thanks to Luke

s=>[...a='abcdefghijklmnopqrstuvwxyz'].filter((c,i)=>eval(s.replace(/\w+/g,s=>[...s].reduce((p,c)=>p|1<<a.search(c),0)))&1<<i).join``

Test cases

let f =

s=>[...a='abcdefghijklmnopqrstuvwxyz'].filter((c,i)=>eval(s.replace(/\w+/g,s=>[...s].reduce((p,c)=>p|1<<a.search(c),0)))&1<<i).join``

console.log(f('ab + bd'));  // acd
console.log(f('d - ab'));   // ac
console.log(f('ab * cd'));  // cf

Mathematica, 168 bytes

FixedPoint[StringReplace[x_~~x_:>FromCharacterCode[c@x+1]],Table["a",ToExpression@StringReplace[#,x:LetterCharacter..:>ToString@Tr[2^((c=ToCharacterCode)@x-97)]]]<>""]&

My initial solution (before the post was edited to clarify that the output must be simplified) was 64 bytes shorter:

Table["a",ToExpression@StringReplace[#,x:LetterCharacter..:>ToString@Tr[2^(ToCharacterCode@x-97)]]]<>""

This just modified that solution to work. It's probably shorter to actually use the methods described in the challenge, but I wanted to put this up anyway.

Explanation:

Replaces each sequence of letters with its corresponding integer by character code arithmetic, then converts the resulting string to an expression (which will automatically simplify to an integer), then produces a string of a characters of length equal to that integer, and finally replaces adjacent identical characters with the next character code up until a fixed point is reached.

Perl 5, 95 bytes

94 bytes of code + -p flag.

s/\w/a x 2**(-97+ord$&)/ge;s/(.*)-\1|\+//;/\*/&&($_=$`x length$');1while s/(.)\1/chr 1+ord$1/e

Try it online!

Three steps here:
- s/\w/a x 2**(-97+ord$&)/ge; converts the input into a string of a only.
- s/(.*)-\1|+//;/*/&&($_=$`x length$') will execute the operator (that are very simple on strings of a): + is the concatenation, - means removing from the first part as many a as there are in the second part, and * means duplicating the first part as many times as there are a in the second part.
- 1while s/(.)\1/chr 1+ord$1/e folds the consecutive same letters into the next letter in the alphabet.