#C, 511 488 474 470 463 454
void f(char*a){char*s;int i=-1,c,o,w,b,e=b=w=o=c=1,l=3,n,r,z=i;for(;s=a[++i];c+=s==67,o+=s==79,w+=s==87,b+=s==66,e+=s==69,l+=s==76);r=~-l/2;n=c<o?c:o;n=w<n?w:n;n=b<n?b:n;n=e<n?e:n;n=r<n?r:n;c=c==n;o=o==n;w=w==n;b=b==n;e=e==n;if(l=r==n?l:0)if(l%2)l=2;else l=1,c=o=w=b=e=0;else l+=l%2;n=c+o+w+b+e+l;for(printf("%s",a);s=n?a[++z]:0;s==67&&c?n--,c--:0,s==79&&o?n--,o--:0,s==87&&w?n--,w--:0,s==66&&b?n--,b--:0,s==69&&e?n--,e--:0,s==76&&l?n--,l--:0,putchar(s));}
Readable format + explanation:
void f(char*a){
//a = input
char*s;
int i=-1,c,o,w,b,e=b=w=o=c=1,l=3,n,r,z=i;//c,o,w,b,e all start at 1; L starts at 3
for(;s=a[++i];c+=s==67,o+=s==79,w+=s==87,b+=s==66,e+=s==69,l+=s==76);
//loop to obtain number of times each character C,O,W,B,E,L is found in string (using the ASCII numeric values of each letter)
//to get an extra cowbell we need to increment C,O,W,B,E by 1 and L by 2 (two Ls in cowbell); except we don't have to because we already did that by starting them at c=1, o=1, w=1, b=1, e=1, L=3 when we declared them.
r=~-l/2;
//r is half of (1 less the number of times L is in string (+ init value))
n=c<o?c:o;n=w<n?w:n;n=b<n?b:n;n=e<n?e:n;n=r<n?r:n;
//n is the number of times that the least occouring character appears in the string, (use R instead of L since cowbell has two L's in it and we just need ~-l/2)
c=c==n;o=o==n;w=w==n;b=b==n;e=e==n;
//convert c,o,w,b,e to BOOL of whether or not we need 1 more of that letter to create one more cowbell (logic for L handled below since it's trickier)
if(l=r==n?l:0)//if L-1/2 is [or is tied for] least occurring character do below logic, else set l to 0 and skip to `else`
if(l%2)//if l is divisible by 2 then we need 2 more Ls
l=2;
else //otherwise we just need 1 more l and no other letters
l=1,c=o=w=b=e=0;
else //add 1 to L if it's divisible by 2 (meaning just 1 more L is needed in addition to possibly other C,O,W,B,E letters) (*Note: L count started at 3, so a count of 4 would be divisible by 2 and there is only 1 L in the string)
l+=l%2;
n=c+o+w+b+e+l;
//n = number of specific characters we need before we reach 1 more cowbell
for(printf("%s",a);s=n?a[++z]:0;s==67&&c?n--,c--:0,s==79&&o?n--,o--:0,s==87&&w?n--,w--:0,s==66&&b?n--,b--:0,s==69&&e?n--,e--:0,s==76&&l?n--,l--:0,putchar(s));
//loop starts by printing the original string, then starts printing it again one character at a time until the required number of C,O,W,B,E,L letters are reached, then break (s=n?a[++z]:0) will return 0 when n is 0. Each letter subtracts from n only when it still requires letters of its type (e.g. b?n--,b--:0)
}
Some Fun Tricks Used:
•When checking characters I type 'w' for the char w which is 3 bytes, but for the characters 'c' and 'b' I can type their ASCII values 99 and 98 respectively to save a byte each time. (Edit: Thanks to @Titus I know do this with all COWBELL letters by using uppercase input only which are all 2 bytes numeric ascii values)
•The second defined int (in this case c) is always set as 1, so I can shorten c=1,o=1,w=1,b=1,e=1 to c,o,w,b,e=b=w=o=c=1 (edit fixed typo and now it turns out this isnt shorter)
•r=~-l/2 is r=(l-1)/2 using bitshifts
•a[++i] I'm getting the character at index(i) and iterating the index all at the same time. I just start i at i=-1 instead of i=0 (I do the same with z and start it as z=i to save another byte)