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newReduce[x_Equal, y_] := Reduce[x /. Solve[y == #, Cases[x, _Symbol, {-1}, 1]], #] /. # -> y &@Unique["a"]

a = 2;
newReduce[x + r y + z == d a + 2 u - op/3, x + y + z]
(* x + y + z == 1/3 (6 d - op + 6 u + 3 y - 3 r y) *)

newReduce[x + r /y + z == d /a + 2 u - op/3, x y + y z]
(* x y + y z == 1/6 (-6 r + 3 d y - 2 op y + 12 u y) && y != 0 *)
newReduce[x + r /y + z == d a + 2 u - op/3, x y z]
(* x y z == -(1/3) z (3 r - 6 d y + op y - 6 u y + 3 y z) && y z != 0 *)

newReduce[x_Equal, y_] := Reduce[x /. Solve[y == #, Cases[y, _Symbol, {-1}, 1]], #]
                          /. # -> y &@Unique["a"]

Clear[a]
newReduce[a == b + c + d - e, c + d]
(* c + d == a - b + e *)

a = 2;
newReduce[x + r y + z == d a + 2 u - op/3, x + y + z]
(* x + y + z == 1/3 (6 d - op + 6 u + 3 y - 3 r y) *)

newReduce[x + r /y + z == d /a + 2 u - op/3, x y + y z]
(* x y + y z == 1/6 (-6 r + 3 d y - 2 op y + 12 u y) && y != 0 *) 

newReduce[x + r /y + z == d a + 2 u - op/3, x y z]
(* x y z == -(1/3) z (3 r - 6 d y + op y - 6 u y + 3 y z) && y z != 0 *)

The combination of Reduce and ReplaceAll may help:

Reduce[a == b + c + d - e /. c -> aa - d, aa] /. aa -> c + d
(* c + d == a - b + e *)

There's a defection in the solution above: it won't work correctly if the original equation contains the variable aa, of course in most cases we can avoid the duplication of name manually without much effort, but here I'd like to add a safer one:

Reduce[a == b + c + d - e /. c -> # - d, #] /. # -> c + d &@Unique["a"]
(* c + d == a - b + e *)

Unique is used to create a symbol that has never been used. This solution will work even if a has got an value:

a = 1;
Reduce[a == b + c + d - e /. c -> # - d, #] /. # -> c + d &@Unique["a"]
(* c + d == 1 - b + e *)

Let's make it into a funtion:

newReduce[x_Equal, y_Plus] := Reduce[x /. First@y -> # - Rest@y, #] /. # -> y &@Unique["a"]

The first argument of newReduce is the equation, and the second one is the "multiple variables", for example:

a = 2;
newReduce[x + r y + z == d a + 2 u - op/3, x + y + z]
(* x + y + z == 1/3 (6 d - op + 6 u + 3 y - 3 r y) *)

The combination of Reduce and ReplaceAll may help:

Reduce[a == b + c + d - e /. c -> aa - d, aa] /. aa -> c + d
(* c + d == a - b + e *)

There's a defection in the solution above: it won't work correctly if the original equation contains the variable aa, of course in most cases we can avoid the duplication of name manually without much effort, but here I'd like to add a safer one:

Reduce[a == b + c + d - e /. c -> # - d, #] /. # -> c + d &@Unique["a"]
(* c + d == a - b + e *)

Unique is used to create a symbol that has never been used. This solution will work even if a has got an value:

a = 1;
Reduce[a == b + c + d - e /. c -> # - d, #] /. # -> c + d &@Unique["a"]
(* c + d == 1 - b + e *)

Let's make it into a funtion:

newReduce[x_Equal, y_] := Reduce[x /. Solve[y == #, Cases[x, _Symbol, {-1}, 1]], #] /. # -> y &@Unique["a"]

The first argument of newReduce is the equation, and the second one is the "multiple variables", for example:

a = 2;
newReduce[x + r y + z == d a + 2 u - op/3, x + y + z]
(* x + y + z == 1/3 (6 d - op + 6 u + 3 y - 3 r y) *)

newReduce[x + r /y + z == d /a + 2 u - op/3, x y + y z]
(* x y + y z == 1/6 (-6 r + 3 d y - 2 op y + 12 u y) && y != 0 *)
newReduce[x + r /y + z == d a + 2 u - op/3, x y z]
(* x y z == -(1/3) z (3 r - 6 d y + op y - 6 u y + 3 y z) && y z != 0 *)

As you see, now it works not only for the "multiple variables" with the head Plus.

The combination of Reduce and ReplaceAll may help:

Reduce[a == b + c + d - e /. c -> aa - d, aa] /. aa -> c + d
(* c + d == a - b + e *)

There's a defection in the solution above: it won't work correctly if the original equation contains the variable aa, of course in most cases we can avoid the duplication of name manually without much effort, but here I'd like to add a safer one:

Reduce[a == b + c + d - e /. c -> # - d, #] /. # -> c + d &@Unique["a"]
(* c + d == a - b + e *)

Unique is used to create a symbol that has never been used. This solution will work even if a has got an value:

a = 1;
Reduce[a == b + c + d - e /. c -> # - d, #] /. # -> c + d &@Unique["a"]
(* c + d == 1 - b + e *)

The combination of Reduce and ReplaceAll may help:

Reduce[a == b + c + d - e /. c -> aa - d, aa] /. aa -> c + d
(* c + d == a - b + e *)

There's a defection in the solution above: it won't work correctly if the original equation contains the variable aa, of course in most cases we can avoid the duplication of name manually without much effort, but here I'd like to add a safer one:

Reduce[a == b + c + d - e /. c -> # - d, #] /. # -> c + d &@Unique["a"]
(* c + d == a - b + e *)

Unique is used to create a symbol that has never been used. This solution will work even if a has got an value:

a = 1;
Reduce[a == b + c + d - e /. c -> # - d, #] /. # -> c + d &@Unique["a"]
(* c + d == 1 - b + e *)

Let's make it into a funtion:

newReduce[x_Equal, y_Plus] := Reduce[x /. First@y -> # - Rest@y, #] /. # -> y &@Unique["a"]

The first argument of newReduce is the equation, and the second one is the "multiple variables", for example:

a = 2;
newReduce[x + r y + z == d a + 2 u - op/3, x + y + z]
(* x + y + z == 1/3 (6 d - op + 6 u + 3 y - 3 r y) *)