Why is it that when proving trig identities, one must work both sides independently?

Why can't I manipulate the entire equation?

You can. The analytical method for proving an identity consists of starting with the identity you want to prove, in the present case $$ \begin{equation} \frac{\sin \theta -\sin ^{3}\theta }{\cos ^{2}\theta }=\sin \theta,\qquad \cos \theta \neq 0 \tag{1} \end{equation} $$ and establish a sequence of identities so that each one is a consequence of the next one. For the identity $(1)$ to be true is enough that the following holds $$ \begin{equation} \sin \theta -\sin ^{3}\theta =\sin \theta \cos ^{2}\theta \tag{2} \end{equation} $$ or this equivalent one $$ \begin{equation} \sin \theta \left( 1-\sin ^{2}\theta \right) =\sin \theta \cos ^{2}\theta \tag{3} \end{equation} $$ or finally this last one $$ \begin{equation} \sin \theta \cos ^{2}\theta =\sin \theta \cos ^{2}\theta \tag{4} \end{equation} $$

Since $(4)$ is true so is $(1)$.

The book indicated below illustrates this method with the following identity $$ \frac{1+\sin a}{\cos a}=\frac{\cos a}{1-\sin a}\qquad a\neq (2k+1)\frac{\pi }{2} $$

It is enough that the following holds $$ (1+\sin a)(1-\sin a)=\cos a\cos a $$

or $$ 1-\sin ^{2}a=\cos ^{2}a, $$

which is true if $$ 1=\cos ^{2}a+\sin ^{2}a $$

is true. Since this was proven to be true, all the previous indentities hold, and so does the first identity.

Reference: J. Calado, Compêndio de Trigonometria, Empresa Literária Fluminense, Lisbon, pp. 90-91, 1967.

You've got a pretty good handle on the situation. It's not so much that you can't manipulate the potential identity as an equation as that, in general, most people shouldn't manipulate the potential identity as an equation. The key part is what you said—use the manipulation to arrive at a true statement (that's your scratch-work), then work backwards to write your proof: starting with a true statement and arriving at the identity.

In your last example, since $\cos\theta$ is in the denominator, $\theta=\frac{\pi}{2}$ would not be in the domain of the identity, so it's okay to simplify to $\sin\theta$.