Ignoring Column #1, Column #7, Row #2, Row #7, we can complete the rook polynomial for the smaller board (Since machine 1 has to process job 2, and machine 7 has to process job 7). This gives us the rook polynomial of
$R(x,B) = 1 + 4x + 6x^2 + 4x^3 + x^4$
If we let $A_i$ be that a machine is assigned to the shaded area, by principle of inclusion-exclusion we can arrive to the following:
$7! - (6!4 - 5!6 + 4!4 - 3!1)$
Would this be the right approach to the problem?
You have the right basic approach, but you’ve not carried it out correctly. After you remove rows $2$ and $7$ and columns $1$ and $7$, you have this board, where I’ve numbered the forbidden squares:
$$\begin{array}{|c|c|c|} \hline &&&&\;\;\\ \hline &&&4&\\ \hline &&3&&\\ \hline &2&&&\\ \hline 1&&&&\\ \hline \end{array}$$
The deleted rows and columns do not contribute anything, since the assigments there are fixed.
The rook polynomial for the forbidden sub-board is $(1+x)^4=1+4x+6x^2+4x^3+x^4$. If $A_k$ is the set of placements of $5$ rooks on this board that have a rook on cell $k$ (for $k=1,\ldots,4$), then an inclusion-exclusion calculation shows that the number of ‘bad’ job-machine assignments is
$$\begin{align*} \left|\bigcup_{k=1}^4A_k\right|&=\sum_{\varnothing\ne I\subseteq[4]}(-1)^{|I|-1}\left|\bigcap_{k\in I}A_k\right|\\ &=\sum_{\varnothing\ne I\subseteq[4]}(-1)^{|I|-1}(5-|I|)!\\ &=\sum_{i=1}^4\binom4i(-1)^{i-1}(5-i)!\;. \end{align*}$$
The number of acceptable assignments is therefore
$$\begin{align*} 5!-\sum_{i=1}^4\binom4i(-1)^{i-1}(5-i)!&=5!+\sum_{i=1}^4\binom4i(-1)^i(5-i)!\\ &=\sum_{i=0}^4\binom4i(-1)^i(5-i)!\\ &=5!-4\cdot4!+6\cdot3!-4\cdot2!+1\cdot1!\\ &=53\;. \end{align*}$$
In the language of rook placements, we really are placing only $5$ rooks, not $7$, because $2$ of the original $7$ have forced placements and so contribute no variety.