In all of these problems, you are taking a structure with operations that you "already know", and are trying to test to see whether a new operation, defined in terms of the ones you know, satisfies certain properties.
It's usually best to denote the "new" operation with a symbol that is unlikely to generate confusion. You've been doing this elsewhere, but here you are using $\cdot$, which can easily be confused with regular integer multiplication. So I'm going to replace it with $\odot$, if you don't mind.
Let's approach it systematically, and essentially "following your nose." I'm going to check a few things that you seem to take for granted, just to get familiarity with the operation.
Now, we have the set $T=\mathbb{Z}\times\mathbb{Q}$, and the operation $(a,b)\odot (c,d) = (-ac, b+d+2).$ First, the operation is commutative, since integer multiplication and addition of rationals is commutative. It is also associative: $\begin{align*} \Bigl( (a,b)\odot (c,d)\Bigr)\odot (x,y) &= (-ac, b+d+2)\odot (x,y)\\ &=(-(-ac)x,(b+d+2)+y+2)\\ &= (acx, b+d+y+4)\\ (a,b)\odot \Bigl((c,d)\odot (x,y)\Bigr)&= (a,b)\odot (-cx, d+y+2)\\ &= (-a(-cx), b+(d+y+2)+2)\\ &= (acx, b+d+y+4). \end{align*}$ So the operation is commutative and associative, and we do indeed have a semigroup.
Now we are trying to see whether this is a monoid (semigroup with identity). So we are trying to see whether there exist $\alpha\in\mathbb{Z}$ and $\beta\in\mathbb{Q}$ such that, for all $a\in\mathbb{Z}$ and $b\in\mathbb{Q}$, we have $(\alpha,\beta)\odot(a,b) = (a,b)$ (exactly as you did). This amounts to solving the equations you solved, which you did correctly, and so we discovered that $(T,\odot)$ is indeed a monoid, and that the identity element is $e_T = (-1,-2)$. Great!
To figure out what elements have inverses, you do set $(a,b)\odot (x,y) =e_T = (-1,-2)$, and figure out if you can determine necessary and sufficient conditions on $a$ and $b$ for $x$ and $y$ to exist. As you note, we have: $(-1,-2) = (a,b)\odot(x,y) = (-ax, b+y+2).$ This requires $-ax = 1$, which means that $a$ has to be an integer that has a multiplicative inverse in the integers (namely, $-x$); there are only two possibilities: $a=1$ or $a=-1$.
On the other hand, for any rational number $b$ we can always find a rational number $y$ such that $b+y+2=-2$, namely, as you note, $y = -(b+4)$. So it would seem that provided that $a=\pm 1$, then $(a,b)$ has an $\odot$-inverse in $T$. And, indeed, if $a=\pm 1$, then we have $(\pm 1,b)\odot (\pm1,-b-4) = (-(\pm1)(\pm 1), b-b-4+2) = (-1-2).$ So a necessary and sufficient condition for $(a,b)\in T$ to be $\odot$-invertible is that $a=\pm 1$.