@Carl Love Since a＝b cannot become or form a cycle (∵ b＝a is missing or lacking), the output should exclude [a, b].

@Carl Love Surely need not necessarily be a permutation; They are just a list of "ordered pairs". (However, sometimes such a list does perform a permutation.)

Remark. Something like this: the rules (instead of equalities) g＝d, c＝g, d＝c construct a cycle g→d→c→g, so the result includes [g, d, c] (and for the same reason, the result includes [f, d, c] as well).

@dharr & @acer Thanks. It appears that nothing returns when I just give simple bounds on variables: 
 

Download anExplicitEnumeration.mw

@Kitonum I find that to obtain the Sol, this is enough: 

identify(solve(eqs =~ 0.));
 = 
 /         14                1        1        1        1        
{ k[1] = - -- k[10] - k[9] - - k[7] + - k[4] + - k[8] - - k[5], k
 \         3                 3        3        3        3        

          155                   10        7        13     
  [2] = - --- k[10] - 11 k[9] - -- k[7] + - k[4] + -- k[8]
           3                    3         3        3      

     4       
   - - k[5], 
     3       

         44                  4        16        4        1       
  k[3] = -- k[10] + 3 k[9] + - k[7] - -- k[4] - - k[8] + - k[5], 
         3                   3        3         3        3       

  k[4] = k[4], k[5] = k[5], 

         122                  4        5        13        1       
  k[6] = --- k[10] + 8 k[9] + - k[7] + - k[4] - -- k[8] + - k[5], 
          3                   3        3        3         3       

                                                      \ 
  k[7] = k[7], k[8] = k[8], k[9] = k[9], k[10] = k[10] }
                                                      / 

As @acer says, using a finite working precision may lead to less correct answers, but this way is feasible after all. Thanks.

@dharr Sorry for late reply. I think that the third form (i.e., c) is more suitable for representing the generic solutions. Many thanks.

@dharr @Axel Vogt Thanks. Is it possible to convert it into the following more ”symmetric“ form? 

[RootOf(_Z^3 - 3*_Z^2 - 10*_Z - 1, index = 3), -4/5*RootOf(_Z^3 - 3*_Z^2 - 10*_Z - 1, index = 3)^2 + 19/5*RootOf(_Z^3 - 3*_Z^2 - 10*_Z - 1, index = 3) + 3/5, 1]:
 = 
[RootOf(_Z^3 - 3*_Z^2 - 10*_Z - 1, index = 3), RootOf(_Z^3 + 10*_Z^2 + 3*_Z - 1, index = 3), 1]:
 = 
[RootOf(_Z^3 - 4*_Z^2 + _Z + 1, index = 1)/RootOf(_Z^3 - 4*_Z^2 + _Z + 1, index = 3), RootOf(_Z^3 - 4*_Z^2 + _Z + 1, index = 2)/RootOf(_Z^3 - 4*_Z^2 + _Z + 1, index = 3), 1]:

@C_R @dharr Thanks for your proposals.

@C_R Thanks. This approach may fail in the absence of ample prior knowledge. Be that as it may, this, after all, is one way of doing it.

@dharr For instance, what about 

expr1 := a^3*b^3*(12*(a*b+a+b)-28*(a+b+1)^2)+4*a^2*b^2*(a+b+1)*(a^2+b^2+(a+1)*(b+1))*(13*(a+b+1)^2-8*(a*b+a+b))+a*b*(7*(a+b+1)^8-40*(a*b+a+b)*(a+b+1)^6+32*(a*b+a+b)^2*(a+b+1)^4+28*(a*b+a+b)^3*(a+b+1)^2-20*(a*b+a+b)^4)+(a+b+1)*(a*b+a+b)*((a+b+1)^4-6*(a*b+a+b)*(a+b+1)^2+6*(a*b+a+b)^2)^2: # %assuming nonnegative

and

expr2 := (a+b+1)^6-27*(a*b+a+b)*(a+b+1)^4+472*(a*b+a+b)^2*(a+b+1)^2+112*(a-1)*(b-1)*(a-b)*(a*b+a+b)*(a+b+1)-16*a*b*(2*(a+b+1)^2+3*(a*b+a+b))*(a+b+1)-1152*(a*b+a+b)^3: # %assuming real

? The former one is over nonnegative reals, and the latter one is over the real numbers.

Unexpectedly, Maple's solve (or RealDomain:-solve) is rather slow in solving expr₁＝0. (For the second equation, can Maple simplify or normalize the values?)

@C_R Sorry for the confusion. I don't signify there's a connection between these two questions. I just cite a worksheet that is not created by myself: The purport of it is to show that the problem is irrelevant to the worksheet (by me or by other users).

@C_R As Options Dialog - Export Tab - Maple Help (maplesoft.com) says, there are two different settings. However, this doesn't change the situation. Here is the worksheet: https://www.mapleprimes.com/questions/235749-Better-Way-To-Remove-Entries-In-Piecewise#answer292495.

Export using text for searchable text and smaller file size: remove_undefined１.pdf.
Export using shapes for greater fidelity: remove_undefined２.pdf.

E.g., confirmed.pdf

Download confirmed.mw

@Axel Vogt Well, this is hard to say, since I don't check the results (using other CAS’s). But I tend to believe that it's complete.

@Hullzie16 The key is that the domains of variables have been restricted.

Here is a simple example (just for fun): 

SolveTools:-SemiAlgebraic({36*a^2*b^2-9*a*b*((a+b)^2+7)+(a+b)*((a+b)*((a+b-5)*(a+b)+24)-5)+1=0}):

In the field of real numbers, how many possible solutions exist exactly? 
If you thought ∞, you’re wrong. (But this fact is, in effect, not easy to clarify.)

Can you give the values of k and w?