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. 

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.
