Is $G= \mathbb{Z}$ a group with binary operation defined as $a \cdot b \equiv a - b$? 
Is $G= \mathbb{Z}$  a group with binary operation defined as $a \cdot b \equiv a - b$?

I am pretty sure the answer is NO but I would like some verification on my reasoning below: 
Associativity fails as in:
Let $a,b,c \in G$ then $a \cdot (b \cdot c) = a \cdot (b-c) = a - (b-c) = a - b + c$.
But $(a \cdot b) \cdot c = (a-b) \cdot c = a - b - c \neq a \cdot (b \cdot c)$.
Identity
Let $a \in G$. Then $a \cdot 0 = a - 0 = a$, so $0$ is the additive identity of $a$, hence the identity element in $G$. But $0 \cdot a = 0 - a = -a \neq a \cdot 0$
Inverse
Let $a \in G \backslash \{0\}$. Assume $G$ is a group, then $a$ has an inverse $a^{-1}$. So $a \cdot a^{-1} = a - a^{-1} = a - (-a) \neq 0$.
I am confident about associativity, but I am not sure if my arguments about the inverse and identity are justified. 
 A: The lack of associativity and identity is correctly proven, good job on that note.
You cannot even talk about the inverse, because you need the identity to talk about the inverse (for every $a$ you need $a^{-1}$ so that $a \cdot a^{-1}$ is the identity , but you've already shown there isn't any : when you talk about $a+ a^{-1} \neq 0$, then why are you using $0$ in the paragraph when it is not the identity?).
As pointed out earlier, to prove that something is not a group it is sufficient to contradict any one of the properties. For example, if there's no identity then there's no need to look at associativity. 
The notion of inverse comes only once an identity element has been discovered : if that itself is not there, then we cannot even discuss the inverse axiom.
A: You don't need to disprove every property for $(\mathbb Z,-)$ not to be a group, it suffices to show that at least one property does not hold.
Indeed, as you rightly pointed out, associativity does not work, so it is not a group.
If I tell you that every cat has whiskers, a tail and four paws, it suffices for you to show me that a human doesn't have whiskers in order to deduce that a human is not a cat (you don't have to bother with tail and paws). 
A: Let me make a minor comment about the first part. Just because two expressions look different does not mean they are different, necessarily. So just noting that the expression for 
$a(bc)$ (namely $a−b+c$) seems different form the expression for $(ab)c$ (namely, $a-b-c$) does not justify the assertion that they are unequal. 
In fact, for some values of $a$, $b$, and $c$, they are equal! (for instance, whenever $c=0$). 
So you should pick specific values for $a$, $b$, and $c$, and show explicitly that with those values, $(ab)c$ does not equal $a(bc)$.  The calculations you’ve made should help you pick those values.
Exhibiting even a single specific instance in which the two expressions are different is enough to show the operation is not associative. But what you have right now is not, in my opinion, sufficient; and it’s technically incorrect, since the inequality $a-b-c\neq a-b+c$ does not always hold, but your statement (by omission of quantifiers) is that they are always different.
