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Python 204 202 195 Bytes

from numpy import*
def f(A):
 r,c=A.shape
 z,s=zeros((r,1)),array([0,2,c+3])
 B=hstack((z,A,z)).flat
 for i in range(1,(r-1)*(c+2)):
    if B[i]and not any(B[s]):return 1<0
    s+=1
 return any(B[i:])

Expects A to be a 2D numpy array.

Takes the matrix, pads zero columns left and right and flattens the matrix. s is a stencil that points to the left, right and lower element. The loop checks each element except the last line if it is 1 and at least one of its stencil is 1, returns False otherwise. Afterwards, check if the last line contains any 1.

Two testcases for you:

I1 = '001001\n011001\n010111\n110101\n011101\n001011'
A1 = array([int(c) for c in I1.replace('\n','')]).reshape(6,6)
print f(A1) #True

I2 = '001100\n111111\n110101\n010011\n111011'
A2 = array([int(c) for c in I2.replace('\n','')]).reshape(5,6)
print f(A2) #False

Edit1: 1<0 is shorter than False

Edit2: flat is a fine alternative to flatten() and using tabulators for the second intendation in the loop

Python + NumPy 204 202 195 Bytes

from numpy import*
def f(A):
 r,c=A.shape
 z,s=zeros((r,1)),array([0,2,c+3])
 B=hstack((z,A,z)).flat
 for i in range(1,(r-1)*(c+2)):
    if B[i]and not any(B[s]):return 1<0
    s+=1
 return any(B[i:])

Expects A to be a 2D numpy array.

Takes the matrix, pads zero columns left and right and flattens the matrix. s is a stencil that points to the left, right and lower element. The loop checks each element except the last line if it is 1 and at least one of its stencil is 1, returns False otherwise. Afterwards, check if the last line contains any 1.

Two testcases for you:

I1 = '001001\n011001\n010111\n110101\n011101\n001011'
A1 = array([int(c) for c in I1.replace('\n','')]).reshape(6,6)
print f(A1) #True

I2 = '001100\n111111\n110101\n010011\n111011'
A2 = array([int(c) for c in I2.replace('\n','')]).reshape(5,6)
print f(A2) #False

Edit1: 1<0 is shorter than False

Edit2: flat is a fine alternative to flatten() and using tabulators for the second intendation in the loop

Python 204 202 Bytes

from numpy import*
def f(A):
 r,c=A.shape
 z,s=zeros((r,1)),array([0,2,c+3])
 B=hstack((z,A,z)).flatten()
 for i in range(1,(r-1)*(c+2)):
  if B[i]and not any(B[s]):return 1<0
  s+=1
 return any(B[i:])

Expects A to be a 2D numpy array.

Takes the matrix, pads zero columns left and right and flattens the matrix. s is a stencil that points to the left, right and lower element. The loop checks each element except the last line if it is 1 and at least one of its stencil is 1, returns False otherwise. Afterwards, check if the last line contains any 1.

Two testcases for you:

I1 = '001001\n011001\n010111\n110101\n011101\n001011'
A1 = array([int(c) for c in I1.replace('\n','')]).reshape(6,6)
print f(A1) #True

I2 = '001100\n111111\n110101\n010011\n111011'
A2 = array([int(c) for c in I2.replace('\n','')]).reshape(5,6)
print f(A2) #False

Edit1: 1<0 is shorter than False

Python 204 202 195 Bytes

from numpy import*
def f(A):
 r,c=A.shape
 z,s=zeros((r,1)),array([0,2,c+3])
 B=hstack((z,A,z)).flat
 for i in range(1,(r-1)*(c+2)):
    if B[i]and not any(B[s]):return 1<0
    s+=1
 return any(B[i:])

Expects A to be a 2D numpy array.

Takes the matrix, pads zero columns left and right and flattens the matrix. s is a stencil that points to the left, right and lower element. The loop checks each element except the last line if it is 1 and at least one of its stencil is 1, returns False otherwise. Afterwards, check if the last line contains any 1.

Two testcases for you:

I1 = '001001\n011001\n010111\n110101\n011101\n001011'
A1 = array([int(c) for c in I1.replace('\n','')]).reshape(6,6)
print f(A1) #True

I2 = '001100\n111111\n110101\n010011\n111011'
A2 = array([int(c) for c in I2.replace('\n','')]).reshape(5,6)
print f(A2) #False

Edit1: 1<0 is shorter than False

Edit2: flat is a fine alternative to flatten() and using tabulators for the second intendation in the loop

Python 204 Bytes

from numpy import*
def f(A):
 r,c=A.shape
 z,s=zeros((r,1)),array([0,2,c+3])
 B=hstack((z,A,z)).flatten()
 for i in range(1,(r-1)*(c+2)):
  if B[i]and not any(B[s]):return False
  s+=1
 return any(B[i:])

Expects A to be a 2D numpy array.

Takes the matrix, pads zero columns left and right and flattens the matrix. s is a stencil that points to the left, right and lower element. The loop checks each element except the last line if it is 1 and at least one of its stencil is 1, returns False otherwise. Afterwards, check if the last line contains any 1.

Two testcases for you:

I1 = '001001\n011001\n010111\n110101\n011101\n001011'
A1 = array([int(c) for c in I1.replace('\n','')]).reshape(6,6)
print f(A1) #True

I2 = '001100\n111111\n110101\n010011\n111011'
A2 = array([int(c) for c in I2.replace('\n','')]).reshape(5,6)
print f(A2) #False

Python 204 202 Bytes

from numpy import*
def f(A):
 r,c=A.shape
 z,s=zeros((r,1)),array([0,2,c+3])
 B=hstack((z,A,z)).flatten()
 for i in range(1,(r-1)*(c+2)):
  if B[i]and not any(B[s]):return 1<0
  s+=1
 return any(B[i:])

Expects A to be a 2D numpy array.

Takes the matrix, pads zero columns left and right and flattens the matrix. s is a stencil that points to the left, right and lower element. The loop checks each element except the last line if it is 1 and at least one of its stencil is 1, returns False otherwise. Afterwards, check if the last line contains any 1.

Two testcases for you:

I1 = '001001\n011001\n010111\n110101\n011101\n001011'
A1 = array([int(c) for c in I1.replace('\n','')]).reshape(6,6)
print f(A1) #True

I2 = '001100\n111111\n110101\n010011\n111011'
A2 = array([int(c) for c in I2.replace('\n','')]).reshape(5,6)
print f(A2) #False

Edit1: 1<0 is shorter than False