Here is a more robust version of what I was doing before. Consider your expression,

This will break up a term into the largest number of fractions, without repeating denominators. Here it is in functional form,

fractionExpand[expr_] := 
 Together@*Plus @@@ GatherBy[List @@ Expand[expr], Denominator] // 
  Total

Here is another, longer test,

expr2 = (1 + m)/b + (1 + n)/a + (1 + o)/c + (d + e + f g)/(
   a b) + (h + i + j k)/(b c) + (l + p q)/(a b c) // Together
(* (a b + a c + b c + c d + c e + c f g + a h + a i + a j k + l + a c m + 
 b c n + a b o + p q)/(a b c) *)

fractionExpand@expr2
(* (d + e + f g)/(a b) + (h + i + j k)/(b c) + (1 + m)/b + (
 1 + n)/a + (1 + o)/c + (l + p q)/(a b c) *)

You get back the original input.

Consider your expression,

But as OP points out, this will not combine terms where the denominator is the same except for a constant factor, like a/(2*b) and c/b. This function should be able to simplify an rational expression like desired in the OP (if you come up with a way to break the function, please let me know).

fractionExpand[expr_] :=
  Replace[Expand@expr, 
  expr2_Plus :> (Together@*Plus @@@ 
      GatherBy[List @@ expr2, Variables@*Denominator] // Total)]

This will break up a term into the largest number of fractions, without repeating denominators. Here are a few tests,

expr2 = (1 + m)/b + (1 + n)/a + (1 + o)/c + (d + e + f g)/(
   a b) + (h + i + j k)/(b c) + (l + p q)/(a b c) // Together

fractionExpand@expr2

You get back the original input.

expr3 = a/(2 b) + c/b + d/e + f/(g + z) + (a - c)/(l m b) // Together

fractionExpand@expr3

expr = (a + m a + b + n b + c + k c + d + e)/(a b);

vars = {a, b};

Haven't tested the versatility of this method yet,

Fold[List @@ Apart[#1, #2] &, expr, vars] // Total
(* (c + d + e + c k)/(a b) + (1 + m)/b + (1 + n)/a *)

Here is a more robust version of what I was doing before. Consider your expression,

expr = (a + m a + b + n b + c + k c + d + e)/(a b);

This gets us almost where we want to go,

Expand@expr
(* 1/a + 1/b + c/(a b) + d/(a b) + e/(a b) + (c k)/(
 a b) + m/b + n/a *)

But we have too many terms with the same denominator, so we can use GatherBy to group them, then simplify the sums of terms with the same denominator, then sum it all back up,

Together@*Plus @@@ GatherBy[List @@ Expand[expr], Denominator] // Total
(* (c + d + e + c k)/(a b) + (1 + m)/b + (1 + n)/a *)

This will break up a term into the largest number of fractions, without repeating denominators. Here it is in functional form,

fractionExpand[expr_] := 
 Together@*Plus @@@ GatherBy[List @@ Expand[expr], Denominator] // 
  Total

Here is another, longer test,

expr2 = (1 + m)/b + (1 + n)/a + (1 + o)/c + (d + e + f g)/(
   a b) + (h + i + j k)/(b c) + (l + p q)/(a b c) // Together
(* (a b + a c + b c + c d + c e + c f g + a h + a i + a j k + l + a c m + 
 b c n + a b o + p q)/(a b c) *)

fractionExpand@expr2
(* (d + e + f g)/(a b) + (h + i + j k)/(b c) + (1 + m)/b + (
 1 + n)/a + (1 + o)/c + (l + p q)/(a b c) *)

You get back the original input.