How can AES-like block cipher without non-linear component be broken?
Since the sbox is affine, you can view $s(v)=mv+b$ where $m$ is a 8-by-8 matrix and $b$ and $v$ is a dimension 8 vector over $F_2$. I will show you how  to find $m$ and $b$ in Pari-GP. If it is not clear how to use that combined with the description of AES to create an even bigger linear equation and solve, ask.

Write $\mathbf{0} =(0, 0, 0, 0, 0, 0, 0, 0)$.
Note that $s(\mathbf{0}) = b$, so all that is left to do is find
$m$. Define
$$l(v) = s(v) + s(\mathbf{0}) = mv + b + m\mathbf{0}+b=mv + b + b = mv$$
Note this is a linear function:

s(n) = sbox[n+1] \\ pari-gp is 1-indexed
l(n) = bitxor(s(n), s(0))
a = 10
b = 37
print(  bitxor(l(a),l(b)) == l(bitxor(a,b))  )


will output 1 (True in Pari-GP). To find out what the matrix is,

m = matrix(8, 8, {j}, {i}, {
    ei = 1 &lt;&lt; (i-1);
    ej = 1 &lt;&lt; (j-1);
    bitand(s(ei), ej) == ej
})

print(m)


which outputs

[1, 0, 0, 0, 0, 1, 0, 1;
 0, 0, 0, 1, 1, 0, 0, 1;
 1, 1, 0, 1, 1, 0, 1, 0;
 0, 0, 1, 0, 0, 1, 1, 0;
 0, 1, 1, 0, 0, 1, 1, 1;
 1, 1, 1, 0, 1, 0, 0, 1;
 0, 0, 1, 1, 0, 1, 1, 0;
 0, 0, 1, 0, 0, 1, 0, 1]


Note that because of how the indices work it is a little weird to read. The 128-th entry in your sbox, which corresponds to the vector $(0, 0, 0, 0, 0, 0, 0, 1)^T$ is 179, and the last column is $(1, 1, 0, 0, 1, 1, 0, 1)^T$ and binary(179) = [1, 0, 1, 1, 0, 0, 1, 1].