Let $T=\mathbb Z_7 \times \mathbb Z_{10}$ and let $\odot$ be the operation defined as follows:
$$\begin{aligned} (a,b)\odot(c,d) = (2+a+c, 3bd)\end{aligned}$$
Find the identity element, the inverse elements and characterize all the invertible elements for $(T, \odot)$.
Identity element
In order to find the identity element for $T$:
$$\begin{aligned} (a,b)\odot(e,\varepsilon) = (2+a+e, 3b\varepsilon)\end{aligned}$$
which leads to
$$\begin{aligned} 2+a+e=a \Rightarrow2+e=0\Rightarrow e=-2 = 5 \text{ (mod 7) }\end{aligned}$$
$$\begin{aligned} 3b\varepsilon=b \Rightarrow3\varepsilon=1 \Rightarrow \varepsilon=7 \text{ (mod 10) }\end{aligned}$$
hence the identity element is $(5,7)$.
Inverse and invertible elements
It's time to find the inverse elements, therefore the following needs to happen:
$$\begin{aligned} (a,b)\odot(\alpha,\beta) = (5,7)\end{aligned}$$
so
$$\begin{aligned} 2+a+\alpha = 5 \Rightarrow \alpha = 5 -2 -a = 5+5-a=3-a \end{aligned}$$
$$\begin{aligned} 3b\beta = \frac{7}{3}b^{-1} = 9b^{-1} \end{aligned}$$
does it suffice stating that as $10$ is a non-prime number, $\mathbb Z_{10}$ has zero-divisors, so not all the elements are invertible, in particular:
$$\begin{aligned} (\forall (a,b) \in \mathbb Z_7 \times \mathbb Z_{10}) \text{ } \exists b^{-1} \in (\mathbb Z_7 \times \mathbb Z_{10}, \odot) \Leftrightarrow gcd(b,10)=1 \end{aligned}$$
Is my conclusion right? If so, is there any other way (perhaps more elegant) to say just that?
The computation of the identity is incorrectly justified, as has been noted. From $$3b\varepsilon \equiv b \pmod{10}$$ we can conclude that $(3\varepsilon-1)b\equiv 0\pmod{10}$. This must hold for all $b$. If $\gcd(b,10)=1$, then we know $b$ is invertible modulo $10$, so then we can conclude that $3\varepsilon\equiv 1\pmod{10}$ is necessary. This forces $\varepsilon\equiv 7\pmod{10}$. So the only possible identity will be $(5,7)$. Then you need to check that this works. But note that we cannot go directly from $3b\varepsilon b\equiv b\pmod{10}$ to $3\varepsilon\equiv 0\pmod{10}$ when we make no assumptions about $b$.
It's bad form to start using symbols like $b^{-1}$ before you have established that such an object actually exists. Rather, you get to $3b\beta\equiv 7\pmod{10}$. At this point, you make the observation that this requires that $\gcd(b,10)=1$ (since $1=\gcd(7,10)=\gcd(3b\beta,10)$, and $\gcd(b,10)$ divides $\gcd(3b\beta,10)$). So you must have $\gcd(b,10)=1$, which is now a necessary condition. To show it is sufficient, you note that setting $\beta=9r$, where $br\equiv 1\pmod{10}$, will do.
The final formula is incorrectly written, since you are not looking for $b^{-1}\in\mathbb{Z}_7\times\mathbb{Z}_{10}$. Rather, what you want to write is:
$$\forall (a,b)\in\mathbb{Z}_7\times\mathbb{Z}_{10}\Bigl(\exists x\in\mathbb{Z}_7\times\mathbb{Z}_{10}((a,b)\odot x=x\odot(a,b)=(2,5))\iff \gcd(b,10)=1\Bigr).$$ Note that the biconditional goes with the existence, and that what you are looking for is not $b^{-1}$, but rather