Well, mostly true.
The scaling property, $\delta(cx) = \frac{1}{|c|}\delta(x)$, is easily provable from definition.
The derivative of the step function though, $\frac{d}{dx}\theta(x)=\delta(x)$, rather than the $\frac12\delta(x)$ that you have. That one-half factor threw me off at first; Ive never seen that.
That said, its possible different sources have different definitions (maybe ?!). These functions are rather artificial, after all, and especially the Dirac delta. Its definition doesnt really jive with the rest of mathematics. You cant really differentiate the unit step since there is a jump discontinuity, and what does it even mean for the Dirac delta to "equal infinity at a single point"? How do you integrate over that? How do you take $\frac15$ of that? You can do these things because they are built into the definition by shear willpower.
These are just functions that are used in practice because they are useful and practical, rather than because of any rigor they may or may not have. I just sort of embrace it with suspicion when Im applying them. Whatever rule you use though, just be consistent.
In conclusion, though I wouldnt use the $\frac12$ factor in your derivative identity (unless told otherwise), the two expressions you end with do seem to be equal. All of your math follows logically.
Its reasonable too. Why shouldnt their derivatives be equal if they themselves are equal? $\theta(ct-z) = \theta(t-z/c)$, after all. The zero of the two arguments are the same, ergo the function jumps from 0 to 1 at the same point.
