Your examples work nicely. Let $\mathbb{Z}_n$ be the cyclic group of order $n$. The following theorem is useful in looking at this sort of situation (taken from Contemporary Abstract Algebra by Gallian, 5th ed.):
Let $a$ be an element of order $n$ in a group and let $k$ be a positive integer. Then $\langle a^k\rangle=\langle a^{\gcd(n,k)} \rangle$ and $|a^k|=n/\gcd(n,k)$.
So how can we apply this? Well, clearly $1$ is an element of $\mathbb{Z}_n$ of order $n$. Then (now in additive notation to be consistent with the operation in $\mathbb{Z}_n$) $\langle k\cdot 1 \rangle = \langle k \rangle = \langle \gcd(n,k)\cdot 1 \rangle$, and so if $\gcd(n,k)=1$ then $\langle k \rangle = \langle 1 \rangle = \mathbb{Z}_n$.
This also follows by the second part of the theorem, by noting that $|\langle k \rangle |=|k|$. If $\gcd(n,k)=1$ then $|k|=\frac{n}{1}=n=|\langle k \rangle |$ so $k$ generates $\mathbb{Z}_n$. On the other hand, if $m=\gcd(n,k)\ne 1$ then $|k|=\frac{n}{m}=|\langle k \rangle |<n$, so $k$ does not generate $\mathbb{Z}_n$ but rather a subgroup of order $\frac{n}{m}$.
There are two things wrong with your proof as it stands - first two subgroups of the same group can never have an empty intersection - the identity will be an element of both.
Second, there are groups of order $pq$ which are not commutative (the smallest has order $21$, and it happens whenever $p\gt q$ and $q$ is a factor of $p-1$ (so here $3$ is a factor of $7-1=6$), so if you want elements to commute, you need to prove it.
The way to do this is via a rather better counting argument.
The consequence of every element other than the identity having order other $p$ is that each element of order $p$ is in a cyclic subgroup of order $p$ and this subgroup has $p-1$ non-identity elements. These subgroups are all distinct and if there are $n$ of them we can count elements to get $$pq=n(p-1)+1$$ so that $p(n-q)=n-1$
Now we can choose the higher prime for $p$ so that $p\gt q$.
It is easy to see successively $n\gt 1$, $n\gt p$ (because $p$ is a factor of $n-1$) but this means also that $pq\gt p^2-p+1$ so that $q\gt p-1$ and this leads to a contradiction.
Sylow's theorems are essentially counting theorems, using the properties of prime numbers - and there are several other significant counting arguments in group theory.
[Added in response to comments]
Now suppose $p\lt q$. Then we have more cases.
We can have $p=2$ so the order of the group is $2q$ and it is easy to show that if all elements have order $2$ then the group is commutative and if finite has order a power of $2$.
If $p\gt 2$ then $n\gt 1$ and therefore $n\gt q$ and $n=mp+1$ for some integer $m\ge 2$.
We then have $q-1=m(p-1)$
And I think we probably have to do something approximating a proof of Sylow to show that this doesn't work, as here the arithmetic runs out.
Note that if $p=q$ we scrape by with $n=p+1$ and $p^2=(p+1)(p-1)+1$
Best Answer
No, you cannot conclude that the order is the least common multiple in this situation.
For an easy example, consider $G=C_2\times C_3$, let $a=(1,1)$ and $b=(0,2)$. Then $\langle b\rangle \subseteq \langle a\rangle$ but they are not equal, $o(a) = 6$, $o(b)=3$, and $ab=(1,0)$ has order $2$. (This is just $C_6$, with $a$ a generator and $b=a^2$).
Added. The problem has been changed to exclude either group being contained in the other. This only requires an easy tweak: take $G=C_2\times C_3\times C_2$, $a=(1,1,0)$, $b=(0,2,1)$. Then $o(a)=6$, $o(b)=6$, $ab=(1,0,1)$ has order $2$, and neither subgroup is contained in the other.
In general, the best you can say is that: $$\begin{align*} \frac{\mathrm{lcm}(o(a),o(b))}{\gcd(o(a),o(b))} &\Bigm| \frac{\mathrm{lcm}(o(a),o(b))}{|\langle a\rangle\cap \langle b\rangle|}\\ \strut\\ \frac{\mathrm{lcm}(o(a),o(b))}{|\langle a\rangle\cap \langle b\rangle|}& \Bigm| o(ab)\\ \strut\\ o(ab) &\Bigm| \mathrm{lcm}(o(a),o(b)). \end{align*}$$
The key here is that you can always express an element of finite order as a product of pairwise commuting elements of prime power order, pairwise coprime. To see this, suppose $x$ has order $ab$ with $\gcd(a,b)=1$. Then there exist integers $r$ and $t$ such that $1 = ar+bt$. Note that $\gcd(r,b)=\gcd(r,t)=\gcd(a,t)=1$. We have that $x = (x^{ar})(x^{bt})$. Now, $x^a$ has order $b$, and $\gcd(r,b)=1$, so $o(x^{ar}) = o((x^a)^r) = b/\gcd(b,r) = b$. Symmetrically, $o(x^{bt}) = a$. So $x$ is a product of an element of order $a$ and an element of order $b$. Lather, rinse, and repeat (or do induction on the number of prime factors) to get that if $x$ has order $n$, and $n=p_1^{a_1}\cdots p_r^{a_r}$ is the prime factorization of $n$, then $x$ can be written as $x=x_1\cdots x_n$ with $o(x_i)=p_i^{a_i}$.
So now think about your $a$ and $b$; primes that only occur in one of $o(a)$ and $o(b)$ will occur in $o(ab)$; and more generally a bit of work shows that if the highest power of $p$ that divides $o(a)$ is different from the highest power of $p$ that divides $o(b)$, then the highest power of $p$ that divides $o(ab)$ is the maximum of the two (giving you the "correct" exponent for the least common multiple). So the key lies in situations where the highest power of $p$ that shows up in $o(a)$ and in $o(b)$ is equal. Clearly, you can try to arrange things so that the $p$-part of $a$ and the $p$-part of $b$ cancel out (or just partially cancel out) but the rest of the parts are disjoint, so that you get $o(ab)$ strictly smaller than $\mathrm{lcm}(o(a),o(b))$, while also ensuring that neither of $\langle a\rangle$ and $\langle b\rangle$ contain each other. That is what I did: the $2$-parts of $a$ and $b$ don't interact, but the $3$-parts cancel each other.