Compactify the query string

JavaScript (ES6),  103 99 87  86 bytes

Saved 1 byte thanks to @MatthewJensen

s=>Object.values(o=/(\w*=)(\w*)/g,s.replace(o,(s,k,v)=>o[k]=o[k]?o[k]+[,v]:s)).join`&`

Try it online!

Commented

s =>                  // s = query string
  Object.values(      // get the values of ...
    o =               //   ... the object of this regular expression, which is
      /(\w*=)(\w*)/g, //   re-used to store the keys and values of the query
    s.replace(        //   match each key with the '=' sign and the corresponding
      o,              //   value, using the regular expression defined above
      (s, k, v) =>    //   for each matched string s, key k and value v:
        o[k] =        //     update o[k]:
          o[k] ?      //       if it's already defined:
            o[k] +    //         get the current value
            [, v]     //         append a comma, followed by v
          :           //       else:
            s         //         set it to the entire matched string
                      //         (key, '=', first value)
    )                 //   end of replace()
  ).join`&`           // end of Object.values(); join with '&'

JavaScript, 263 201 196 194 190 189 188 173 161 bytes

q=>{w={};q.split`&`.map(x=>{y=x.split`=`;if(!w[y[0]])w[y[0]]=[];w[y[0]].push(y[1])});z=`${Object.entries(w).map(a=>a=[a[0]+'='+a[1].join`,`]).join`&`}`;return z}

Try it online

Python 3.8 (pre-release), 116 bytes

lambda s:(a:=[k.split("=")for k in s.split("&")])and"&".join(b+"="+",".join(d for c,d in a if c==b)for b in dict(a))

Try it online!

-46 bytes thanks to ovs

-1 byte thanks to Jonathan Allan (in Py 3.8 PR, with walrus)