If $a$ is relatively prime to both $m$ and $n$ and if $\gcd(m, n) =1$, I hope to find a formula for $e_{mn} (a)$ in terms of $e_m (a)$ and $e_n (a)$. Note: $e_n (a)$ is the order of $a$ module $n$ and is equal to $\phi(n)$ is a is a primitive root.
$a^{e_n} \equiv 1 \ (\text{mod} \ n) \ \text{and} \ a^{e_{mn}} \equiv 1 \ (\text{mod} \ n) \implies e_n | e_{mn}$. Similarly $e_m | e_{mn}$.
Also, $a^{e_n e_m} \equiv 1 \ (\text{mod} \ n) \ \text{and} \ a^{e_n e_m} \equiv 1 \ (\text{mod} \ m) \implies a^{e_n e_m} \equiv 1 \ (\text{mod} \ mn) e_{mn} | e_n e_m$.
I can't go any further. Obviously $e_m(a) \times e_n(a) = e_{mn}(a)$ if $\gcd(e_n(a), e_m(a))=1$. But how $\gcd(e_n(a), e_m(a))=1$ holds?
We claim $e_{mn}(a)=\mathrm{lcm}(e_m(a),e_n(a)).$ You have already shown that both $e_m(a)$ and $e_n(a)$ divide $e_{mn}(a)$, so there is no smaller possibility. Furthermore, we have
$$a^{\mathrm{lcm}(e_m(a),e_n(a))}=a^{Ke_m(a)}$$
for $K=e_n(a)/\gcd(e_m(a),e_n(a))\in\mathbb{N}$, so
$$a^{\mathrm{lcm}(e_m(a),e_n(a))}\equiv \left(a^{e_m(a)}\right)^K\equiv 1\bmod m.$$
Similarly, it is equivalent to $1\bmod n$, so
$$a^{\mathrm{lcm}(e_m(a),e_n(a))}\equiv 1\bmod mn.$$
We note that this is necessary: If $a=2$, $m=3$, $n=5$, then
$$e_3(2)=2,\ e_5(2)=4,\ e_{15}(2)=4\neq e_3(2)e_5(2).$$