Closure (under multiplication) means: "An integer times an integer is also an integer". It does not mean:
"An integer times something else which results in an integer, means that something else is also an integer."
This is pretty obvious to see:
$2\cdot \frac{1}{2} = 1$.
Here, $2$ is an integer, and the "something else" is $\frac{1}{2}$. Our product is an integer, but it is not the case that we can conclude $\frac{1}{2}$ is an integer; in fact, it is not.
It can be proven that $\Bbb N$ is closed under multiplication (hint: use induction), and using the rules, for $a,b \in \Bbb N$:
$a(-b) = (-a)b = -ab\\
(-a)(-b) = ab$
(If you do not consider $0$ to be a natural number, you have a few more cases to consider, but these are easy)
it is easy to see that if $\Bbb N$ is closed under multiplication, so is $\Bbb Z$.
Perhaps this will be easier to process when you see the differences between rings, and fields. What often throws people off-guard in thinking about this, is that ordinary high-school arithmetic typically takes place in the field of rational numbers, where the non-zero numbers are closed under multiplication, and the equation:
$7x = 3$ has a rational solution, even though it is an integral equation.
You are in the right track, but it is not just the contrapositive of the statement, so perhaps you missed a small detail.
Note that the contrapositive of "$a \not\in H$ implies $a^{-1} \not\in H$" is the following statement:
$a^{-1} \in H$ implies $a \in H$.
With this, use $a = (a^{-1})^{-1}$ to conclude that $H$ is closed under inverses.
Best Answer
Take $G = \Bbb Z$ and $H = \Bbb N$.