$\operatorname{ord}(u)=r,\,\operatorname{ord}(v)=s$ $\,\Rightarrow\,\operatorname{ord}(uv)=rs\,$ if $\,r,s\,$ (co)primes

One can use the same (correct) idea, streamline things a bit, and relax the primality assumptions.

Suppose that $u$ has order $r$ and $v$ has order $s$, where $r$ and $s$ are relatively prime. We show that $uv$ has order $rs$. Since $(uv)^{rs}=1$, it is enough to show that if $k$ is the order of $uv$, then $rs$ divides $k$.

Note that $u^{ks}=u^{ks}v^{ks}=1$. So the order $r$ of $u$ divides $ks$, and since $r$ and $s$ are relatively prime, we conclude that $r$ divides $k$.

Similarly, $s$ divides $k$.

Since $r$ and $s$ are relatively prime, it follows that $rs$ divides $k$.

There is a gap: the prime product $\,rs\,$ has three smaller divisors $\,r,s,\color{#c00}1.\,$ You've ruled out the possibilities $\, k = r,s\,$ but not $\,k = \color{#c00}1.\ $ Below is a quicker way to proceed, where we prove it more generally for any $r,s$ that are coprime, i.e. their gcd $(r,s) = 1.\,$ Let $\,o(x) := {\rm ord}\,x$

$ (uv)^k\! = 1\,\Rightarrow\, u^k\! = v^{-k} =\color{#c00}t \in \langle u\rangle\cap\langle v\rangle\Rightarrow \bigg\lbrace\begin{eqnarray} && {o(t)\mid\ r,s = o(u),\,o(v)},{\rm\ by\ Lagrange}\\ \Rightarrow && o(t)\mid(r,s)\!=\!1\, \Rightarrow\, \color{#c00}{t = 1}\end{eqnarray}$

So $\ (uv)^k\! = 1\iff u^k =\color{#c00}{1} = v^k\iff r,s\mid k\underset{\large (r,s)\,=\,1}\iff rs\mid k,\, $ so $\ rs = o(uv)\ $ by here.

Remark $ $ If Lagrange in unknown, instead: $\,t^{\large r}\! = (u^{\large k})^{\large r}\! = (u^{\large r})^{\large k}\! = 1^{\large k}\! = 1$ thus $\,o(t)\mid r$

To prove that (co)primes $\,r,s\mid k\,\Rightarrow\,rs\mid k\,$ without knowledge of lcm, you can either use the Fundamental Theorem of Arithmetic, or equivalent properties such as Euclid's Lemma, e.g. $\, s\mid k\,\Rightarrow\, k = sn,\,$ so $\,(r,s)=1,\,r\mid sn\,\Rightarrow\, r\mid n,\,$ so $\,n = rm,\,$ so $\,k = sn = s(rm),\,$ so $\,rs\mid k.$ Alternatively, using Bezout's identity, since $\,s,r\,$ are coprime there are integers $\,j,k\,$ with $\,js\!+\!kr = 1\,$ so $\,r\mid sn,rn\,\Rightarrow\,r\mid j(sn)+k(rn) = (js\!+\!kr)n = n,\,$ so the result follows as above. See also this answer for further proofs and comparisons.