Thanks for the tip. It worked:

In[6]:= A = Array[x[#1, #2] &, {3, 3}];
IA = Inverse[A];
B = Array[y[#1, #2] &, {3, 3}];
IB = Inverse[B];
Timing[simplifiedList = {A, B, IA, A, B, IA, A, B, IA, A, B, IA, A, B,
     IA} //. ({x___, y_, z_, t___} /; 
      Simplify[y . z == IdentityMatrix[3]]) :> {x, t};
 simplifiedList == {A, B, B, B, B, B, IA}]

Out[10]= {0.00468, True}

Yes, it does the trick:

In[104]:= A = Array[x[#1, #2] &, {3, 3}];
IA = Inverse[A];
B = Array[y[#1, #2] &, {3, 3}];
IB = Inverse[B];

{A, B, IA, A, B, IA, A, B, IA, A, B, IA, A, B, 
   IA} //. ({x___, y_, z_, t___} /; 
     Simplify[y . z == IdentityMatrix[3]]) :> {x, t};
% == {A, B, B, B, B, B, IA}

(*Timing[ 
{A,B,IA,A,B,IA,A,B,IA,A,B,IA,A,B,IA}//.({x___,y_,z_,t___}/;Simplify[y.\
z==IdentityMatrix[3]]):>{x,t};
%=={A,B,B,B,B,B,IA}
]*)

Out[109]= True

BTW, how to check the performance of the above code block by Timing? I tried the following, but failed:

A = Array[x[#1, #2] &, {3, 3}];
IA = Inverse[A];
B = Array[y[#1, #2] &, {3, 3}];
IB = Inverse[B];

(*{A,B,IA,A,B,IA,A,B,IA,A,B,IA,A,B,IA}//.({x___,y_,z_,t___}/;Simplify[\
y.z==IdentityMatrix[3]]):>{x,t};
%=={A,B,B,B,B,B,IA}*)

Timing[ 
 {A, B, IA, A, B, IA, A, B, IA, A, B, IA, A, B, 
   IA} //. ({x___, y_, z_, t___} /; 
     Simplify[y . z == IdentityMatrix[3]]) :> {x, t};
 % == {A, B, B, B, B, B, IA}
 ]

{0.006172, False}