We have the augmented matrix
$\left[\begin{array}{cccc|c} 1&1&-1&1&1\\ a&1&1&1&b\\ 3&2&0&a&1+a \end{array}\right]\;.$
Subtracting $a$ times the first row from the second row and $3$ times the first row from the third row gives us this matrix:
$\left[\begin{array}{cccc|c} 1&1&-1&1&1\\ 0&1-a&1+a&1-a&b-a\\ 0&-1&3&a-3&a-2 \end{array}\right]\;.$
Interchange the last two rows and multiply the new middle row by $-1$:
$\left[\begin{array}{cccc|c} 1&1&-1&1&1\\ 0&1&-3&3-a&2-a\\ 0&1-a&1+a&1-a&b-a\\ \end{array}\right]\;.$
Subtract $1-a$ times the second row from the third row:
$\left[\begin{array}{cccc|c} 1&1&-1&1&1\\ 0&1&-3&3-a&2-a\\ 0&0&4-2a&-a^2+3a-2&b-a^2+2a-2\\ \end{array}\right]\;.$
If $a\ne 2$, then $4-2a\ne 0$, and we can pivot on $4-2a$ to complete the Gaussian elimination, and the system will be consistent. If $a=2$, the matrix is
$\left[\begin{array}{cccc|c} 1&1&-1&1&1\\ 0&1&-3&1&0\\ 0&0&0&0&b-2\\ \end{array}\right]\;,$
which is consistent if and only if $b=2$.
Thus, the system is consistent when $a\ne 2$, and it is consistent when $a=b=2$.