#Python 2, 297 215 bytes

from itertools import*
T=input()
G=T[0]
print sum(all(T[y][x]in g for x,y in product(g,g))*all(any(T[y][x]==G[0]==T[x][y]for y in g)for x in g)*(G[0]in g)for g in chain(*[combinations(G,n)for n in range(len(G)+1)]))

Try it online

This program works for the example group without ==T[x][y], but I'm still pretty sure it's necessary.

Edit: Now assumes that the identity element of G is always the first.

Ungolfed:

from itertools import*
T=input()
G=T[0]
def f(x,y):return T[y][x]                                           # function
def C(g):return all(f(x,y)in g for x,y in product(g,g))             # closure
def E(g):return[all(f(x,y)==y for y in g)for x in g]                # identity

a=E(G)
e=any(a)
e=G[a.index(1)]if e else-1                                          # e in G

def I(G):return all(any(f(x,y)==e==f(y,x)for y in G)for x in G)     # inverse

#print e
#print C(G),any(E(G)),I(G)

#for g in chain(*[combinations(G,n)for n in range(len(G)+1)]):      # print all subgroups
#   if C(g)and I(g)and e in g:print g

print sum(C(g)*I(g)*(e in g)for g in chain(*[combinations(G,n)for n in range(len(G)+1)]))

Ungolfed TIO

Negative group elements can be supported at no cost by changing -1 to ''.

Python 2, 297 215 bytes

from itertools import*
T=input()
G=T[0]
print sum(all(T[y][x]in g for x,y in product(g,g))*all(any(T[y][x]==G[0]==T[x][y]for y in g)for x in g)*(G[0]in g)for g in chain(*[combinations(G,n)for n in range(len(G)+1)]))

Try it online

This program works for the example group without ==T[x][y], but I'm still pretty sure it's necessary.

Edit: Now assumes that the identity element of G is always the first.

Ungolfed:

from itertools import*
T=input()
G=T[0]
def f(x,y):return T[y][x]                                           # function
def C(g):return all(f(x,y)in g for x,y in product(g,g))             # closure
def E(g):return[all(f(x,y)==y for y in g)for x in g]                # identity

a=E(G)
e=any(a)
e=G[a.index(1)]if e else-1                                          # e in G

def I(G):return all(any(f(x,y)==e==f(y,x)for y in G)for x in G)     # inverse

#print e
#print C(G),any(E(G)),I(G)

#for g in chain(*[combinations(G,n)for n in range(len(G)+1)]):      # print all subgroups
#   if C(g)and I(g)and e in g:print g

print sum(C(g)*I(g)*(e in g)for g in chain(*[combinations(G,n)for n in range(len(G)+1)]))

Ungolfed TIO

Negative group elements can be supported at no cost by changing -1 to ''.

#Python 2, 297 215 bytes

from itertools import*
T=input()
G=T[0]
print sum(all(T[y][x]in g for x,y in product(g,g))*all(any(T[y][x]==G[0]==T[x][y]for y in g)for x in g)*(G[0]in g)for g in chain(*[combinations(G,n)for n in range(len(G)+1)]))

Try it online

This program works for the example group without ==T[x][y], but I'm still pretty sure it's necessary. Negative group elements can be supported at no cost by changing -1 to ''.

Edit: Now assumes that the identity element of G is always the first.

Ungolfed:

from itertools import*
T=input()
G=T[0]
def f(x,y):return T[y][x]                                           # function
def C(g):return all(f(x,y)in g for x,y in product(g,g))             # closure
def E(g):return[all(f(x,y)==y for y in g)for x in g]                # identity

a=E(G)
e=any(a)
e=G[a.index(1)]if e else-1                                          # e in G

def I(G):return all(any(f(x,y)==e==f(y,x)for y in G)for x in G)     # inverse

#print e
#print C(G),any(E(G)),I(G)

#for g in chain(*[combinations(G,n)for n in range(len(G)+1)]):      # print all subgroups
#   if C(g)and I(g)and e in g:print g

print sum(C(g)*I(g)*(e in g)for g in chain(*[combinations(G,n)for n in range(len(G)+1)]))

Ungolfed TIO

#Python 2, 297 215 bytes

from itertools import*
T=input()
G=T[0]
print sum(all(T[y][x]in g for x,y in product(g,g))*all(any(T[y][x]==G[0]==T[x][y]for y in g)for x in g)*(G[0]in g)for g in chain(*[combinations(G,n)for n in range(len(G)+1)]))

Try it online

This program works for the example group without ==T[x][y], but I'm still pretty sure it's necessary.

Edit: Now assumes that the identity element of G is always the first.

Ungolfed:

from itertools import*
T=input()
G=T[0]
def f(x,y):return T[y][x]                                           # function
def C(g):return all(f(x,y)in g for x,y in product(g,g))             # closure
def E(g):return[all(f(x,y)==y for y in g)for x in g]                # identity

a=E(G)
e=any(a)
e=G[a.index(1)]if e else-1                                          # e in G

def I(G):return all(any(f(x,y)==e==f(y,x)for y in G)for x in G)     # inverse

#print e
#print C(G),any(E(G)),I(G)

#for g in chain(*[combinations(G,n)for n in range(len(G)+1)]):      # print all subgroups
#   if C(g)and I(g)and e in g:print g

print sum(C(g)*I(g)*(e in g)for g in chain(*[combinations(G,n)for n in range(len(G)+1)]))

Ungolfed TIO

Negative group elements can be supported at no cost by changing -1 to ''.

#Python 2, 297 bytes

from itertools import*
T=input()
G=T[0]
E=lambda g:[all(T[y][x]==y for y in g)for x in g]
e=G[E(G).index(1)]if any(E(G))else-1
print sum(all(T[y][x]in g for x,y in product(g,g))*all(any(T[y][x]==e==T[x][y]for y in g)for x in g)*(e in g)for g in chain(*[combinations(G,n)for n in range(len(G)+1)]))

Try it online

This program works for the example group without ==T[x][y], but I'm still pretty sure it's necessary. Negative group elements can be supported at no cost by changing -1 to ''.

Ungolfed:

from itertools import*
T=input()
G=T[0]
def f(x,y):return T[y][x]                                           # function
def C(g):return all(f(x,y)in g for x,y in product(g,g))             # closure
def E(g):return[all(f(x,y)==y for y in g)for x in g]                # identity

a=E(G)
e=any(a)
e=G[a.index(1)]if e else-1                                          # e in G

def I(G):return all(any(f(x,y)==e==f(y,x)for y in G)for x in G)     # inverse

#print e
#print C(G),any(E(G)),I(G)

#for g in chain(*[combinations(G,n)for n in range(len(G)+1)]):      # print all subgroups
#   if C(g)and I(g)and e in g:print g

print sum(C(g)*I(g)*(e in g)for g in chain(*[combinations(G,n)for n in range(len(G)+1)]))

Ungolfed TIO

#Python 2, 297 215 bytes

from itertools import*
T=input()
G=T[0]
print sum(all(T[y][x]in g for x,y in product(g,g))*all(any(T[y][x]==G[0]==T[x][y]for y in g)for x in g)*(G[0]in g)for g in chain(*[combinations(G,n)for n in range(len(G)+1)]))

Try it online

This program works for the example group without ==T[x][y], but I'm still pretty sure it's necessary. Negative group elements can be supported at no cost by changing -1 to ''.

Edit: Now assumes that the identity element of G is always the first.

Ungolfed:

from itertools import*
T=input()
G=T[0]
def f(x,y):return T[y][x]                                           # function
def C(g):return all(f(x,y)in g for x,y in product(g,g))             # closure
def E(g):return[all(f(x,y)==y for y in g)for x in g]                # identity

a=E(G)
e=any(a)
e=G[a.index(1)]if e else-1                                          # e in G

def I(G):return all(any(f(x,y)==e==f(y,x)for y in G)for x in G)     # inverse

#print e
#print C(G),any(E(G)),I(G)

#for g in chain(*[combinations(G,n)for n in range(len(G)+1)]):      # print all subgroups
#   if C(g)and I(g)and e in g:print g

print sum(C(g)*I(g)*(e in g)for g in chain(*[combinations(G,n)for n in range(len(G)+1)]))

Ungolfed TIO