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 $a=b=2$.