In general, you're right that $\Phi(AB)\neq\Phi(A)\Phi(B)$--for example, $\Phi(8)=4$, but $\Phi(2)=1$ and $\Phi(4)=2$. At best, we can usually only say that $\Phi(AB)\geq\Phi(A)\Phi(B)$. We *will* have equality when (and only when) $A,B$ are coprime, though.

Let $R_1,R_2$ be rings, and note that $R_1\times R_2$ is also a ring with componentwise addition and multiplication. A given $(x_1,x_2)\in R_1\times R_2$ has a multiplicative inverse if and only if the $x_k$ have multiplicative inverses in $R_k$.

Note that integers $A,B$ are coprime if and only if there exist integers $X,Y$ such that $AX+BY=1$ if and only if $A$ has a multiplicative inverse modulo $B$ and $B$ has one modulo $A$. Thus, there are precisely $\Phi(A)$ elements of $\Bbb Z/A\Bbb Z$ having multiplicative inverse.

By the above discussion, $\Bbb Z/AB\Bbb Z$ will have $\Phi(AB)$ elements with multiplicative inverses, and $(\Bbb Z/A\Bbb Z)\times(\Bbb Z/B\Bbb Z)$ will have $\Phi(A)\Phi(B)$ such elements.

The only thing left to prove is that there is a bijection (in fact, an isomorphism) between $\Bbb Z/AB\Bbb Z$ and $(\Bbb Z/A\Bbb Z)\times(\Bbb Z/B\Bbb Z)$. That $A,B$ are coprime is **crucial** to the proof, so be sure you use it! I'd start with the natural homomorphism $\Bbb Z\to(\Bbb Z/A\Bbb Z)\times(\Bbb Z/B\Bbb Z)$ given by $X\mapsto(X\,mod\, A,X\, mod\, B)$, which readily has kernel $AB\Bbb Z$, and then use the coprimality of $A,B$ to show surjectivity, so the desired conclusion follows by First Isomorphism Theorem.

**Edit**: Your addendum is incorrect. The $\frac{4n}{15}$ lower bound is obtained when we're looking at the first $100$ values of $\Phi(n)$--attained by $30,60,90$, but it is misleading. There are numbers beyond it that fall below that bound, such as $210$. There is in fact *no* lower bounding straight line of positive slope through the origin that applies to all the integers.