$x^4+x^3+x^2+x+1$ irreducible over $\mathbb F_7$

Any element in any field satisfying $x^4+x^3+x^2+x+1=0$ also satisfies $x^5=1$. In other words, it has order $5$ in the multiplicative group of that field.

The field with $p^k$ elements has a multiplicative group with $p^k-1$ elements, so it has elements of order $5$ if and only if $5$ divides $p^k-1$. This is not the case for $p=7$ and $k=1,2,3$, but it is the case for $p=7,k=4$.

Therefore, the smallest field containing $\mathbb{F}_7$ and roots of $x^4+x^3+x^2+x+1$ has $7^4$ elements. If that polynomial were reducible, there would be a smaller splitting field with such roots, but there isn’t.

If you want to continue your approach in case you cannot come up with some clever argument like that in Alon Amit's answer, then you can do it as follows. First, note that $c=1-a$ and $d=b^{-1}$. Thus, the equation $ac+b+d=1$ becomes $$a(1-a)+b+b^{-1}=1\,.\tag{*}$$ The equation $bc+ad=1$ is now $$b(1-a)+ab^{-1}=1\,.$$ Thus, $$a^{-1}\big(1-b(1-a)\big)=b^{-1}=1-b-a(1-a)\,.$$ That is, $$1-b(1-a)=a-ab-a^2(1-a)\,.$$ Consequently, $$(2a-1)b=-1+a-a^2+a^3=(a-1)(a^2+1)\,.\tag{#}$$ Thus, $a\notin\{0,1,4\}$ for the equation above to be possible (noting that $a\neq 0$ and $b\neq 0$). You are left with four choices of $a$, namely, $a\in\{2,3,5,6\}$.

If $a=2$, then, using (#), we get $b=3^{-1}\cdot 5=5\cdot 5=4$, so $b^{-1}=2$, but then $$a(1-a)+b+b^{-1}=2\cdot(-1)+4+2=4\neq 1\,.$$ If $a=3$, then we have by (#) that $b=5^{-1}\cdot20=-3=4$, whence $$a(1-a)+b+b^{-1}=3\cdot(-2)+4+2=0\neq 1\,.$$ If $a=5$, then $b=3^{-1}\cdot (-1)=-5=2$, so $b^{-1}=4$, but then $$a(1-a)+b+b^{-1}=5\cdot(-4)+2+4=0\neq 1\,.$$ Finally, if $a=6$, then $b=4^{-1}\cdot3=2\cdot 3=-1$, whence $b^{-1}=-1$, and $$a(1-a)+b+b^{-1}=6\cdot(-5)+(-1)+(-1)=-4\neq 1\,.$$ Thus, (*) cannot be satisfied.