Have I determined the inverse element for this group correctly? 
The operation $\circ $ has been defined on the set $S=\{ (a,b)|a, b\in \mathbb{R} \}$ in the following manner: $(a,b)\circ (c,d)=(a+c+1,b+d)$. Show that $(S,\circ)$ is a group.

I've already finished showing closure, associativity and identity, but I'm not sure if the inverse element is correct. More specifically, is $(a,b)^{-1}$ in this case, the same as $(a^{-1},b^{-1})$?
Here's what I've tried doing so far:
Let $(a,b)\in S$, then there exists an $(a,b)^{-1}\in S$ such that:
$(a,b)\circ (a,b)^{-1}=(a,b)^{-1}\circ (a,b)=(-1,0)$ (the identity element)
From here, we conclude that:
$a^{-1}=-a-2\\ b^{-1}=b$
And therefore the inverse element of $S$ is $(-a-2,-b)$.
 A: The overall proof method is good, but I'd suggest some changes to wording for clarity and precision.
Writing $(a^{-1}, b^{-1})$ implies that $a$ and $b$ are elements of groups other than $S$. This is just an unnecessary complication, so I'd skip that and stick with $(a, b)^{-1}$. When determining the inverse, instead of variables $a^{-1}$ and $b^{-1}$, just introduce variables $c,d$ and find $c$ and $d$ so that $(a,b) \circ (c,d) = (-1,0)$. (In this case it's actually true that $S$ is a direct product of two groups so that $a$ and $b$ would have their own inverses in those groups, but in similar problems where terms from the two pieces of the pair "mix" we wouldn't have that.)
$(-1,0)$ is the identity element, not "the inverse element".
Though it's fairly obvious for this case, I'd point out directly that the solution you found does in fact plug back in to satisfy both $(a,b) \circ (a,b)^{-1} = (-1,0)$ and $(a,b)^{-1} \circ (a,b) = (-1,0)$. When commutativity is not so obvious, or when the operation is not in fact commutative, this is an important step.
"The inverse element of $S$ is $(-a-2,-b)$" doesn't make sense since $S$ is the whole group, and each element has a different inverse. I'd say just "the inverse of $(a,b)$ [in $S$] is $(-a-2,-b)$". Maybe you were trying to get at "the inverse formula for $S$ is $(-a-2,-b)$" or similar, but that's still not very clear.
A: Some comments:

*

*You ask if $(a,b)^{-1} = (a^{-1}, b^{-1})$. That depends, because you have not said what $a^{-1}$ and $b^{-1}$ are. You might argue "They are the inverse elements of $a$ and $b$!" but that is meaningless by itself - the inverse of an element depends what group it is considered in. And there is no obvious group from which to choose $a$ and $b$.


*You ask whether $(-a-2, -b)$ is the inverse of $S$. This is grammatically incorrect: $S$ is a group, and groups do not have inverses. Group elements have inverses. You have shown "the element $(a,b)$ has inverse $(-a-2, -b)$ in $S$". Or if $S$ is assumed to be the group in question, just "$(a,b)$ has inverse $(-a-2, -b)$".


*You write "$(-1, 0)$ is the inverse element". It is the identity element.


*You separate out $a^{-1}$ and $b^{-1}$, which is unnecessary (and meaningless). The calculation is correct, apart from a typo. All you need is that
$$(a+c+1, b+d) = (-1,0) \implies c = -a-2 \,\, \text{and}\,\, d=-b$$

*

*Your final answer is nevertheless correct.

