Where is the mistake in my 'derivation' of $(uv)' = u'v$?

I find your notation very cumbersome. Now, taking into account that differentiability implies continuity, we can write

$$\frac{f(x)g(x)-f(x_0)g(x_0)}{x-x_0}=\frac{f(x)g(x)-f(x)g(x_0)+f(x)g(x_0)-f(x_0)g(x_0)}{x-x_0}=$$

$$=f(x)\frac{g(x)-g(x_0)}{x-x_0}+\frac{f(x)-f(x_0)}{x-x_0}g(x)\xrightarrow[x\to x_0]{}f(x_0)g'(x_0)+f'(x_0)g(x_0)$$

Complete details. The above is the easiest proof I know of the product rule for derivative. The trick in the first step is also used in general limits.

This has been pointed out in comments already, but the error in the proof is introduced in this step: $$ \lim_{\Delta x\to0} \frac{u(x+\Delta x)v(x+\Delta x)}{\Delta x}-\lim_{\Delta x\to0} \frac{u(x)v(x)}{\Delta x}.$$

Neither of these limits exists (in both cases the quantity goes to infinity), so the difference is indeterminate. Once you start manipulating such a thing as if such limits were valid, it is likely to come out to any number of wrong answers. By inserting a couple more steps afterward, you could reduce the entire expression to zero.

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Taking the idea that a $u'$ has to do with the change in $u$ over the change in $x$, write $\Delta u = u(x+\Delta x) - u(x).$ Likewise write $\Delta v = v(x+\Delta x) - v(x).$

That is, $u(x+\Delta x) = u(x) + \Delta u$ and $v(x+\Delta x) = v(x) + \Delta v.$ Now plug this into your first formula and plod along without knowing where we're going until we get there.

\begin{align} (uv)' &= \lim_{\Delta x\to0} \frac{u(x + \Delta x)v(x + \Delta x) - u(x)v(x)}{\Delta x} \\ &= \lim_{\Delta x\to0} \frac{(u(x) + \Delta u)(v(x) + \Delta v) - u(x)v(x)}{\Delta x} \\ &= \lim_{\Delta x\to0} \frac{(u(x)v(x) + (\Delta u)v(x) + u(x)(\Delta v) + (\Delta u)(\Delta v)) - u(x)v(x)}{\Delta x} \\ &= \lim_{\Delta x\to0} \frac{(\Delta u)v(x) + u(x)(\Delta v) + (\Delta u)(\Delta v)}{\Delta x} \\ \end{align}

Now observe that \begin{align} \lim_{\Delta x\to0} \frac{\Delta u}{\Delta x} &= \lim_{\Delta x\to0} \frac{u(x+\Delta x) - u(x)}{\Delta x} = u'(x), \\ \lim_{\Delta x\to0} \frac{\Delta v}{\Delta x} &= \lim_{\Delta x\to0} \frac{v(x+\Delta x) - v(x)}{\Delta x} = v'(x), \\ \lim_{\Delta x\to0} \frac{(\Delta u)(\Delta v)}{\Delta x} &= \left(\lim_{\Delta x\to0} \frac{\Delta u}{\Delta x}\right) \left(\lim_{\Delta x\to0} \Delta v\right) = 0 \end{align} since $\lim_{\Delta x\to0} \Delta v = 0.$ Putting all of this together,

\begin{align} (uv)' &= v(x)\lim_{\Delta x\to0} \frac{\Delta u}{\Delta x} + u(x)\lim_{\Delta x\to0} \frac{\Delta v}{\Delta x} + \lim_{\Delta x\to0} \frac{(\Delta u)(\Delta v)}{\Delta x} \\ &= v(x) u'(x) + u(x) v'(x) + 0 \\ &= u' v + u v'. \end{align}

No tricks with adding and subtracting a mysterious term or other great inspiration, just distribution of multiplication over addition and beating on the monster until it's dead.

Personally I like the more inspired solutions, but sometimes when you're stuck for inspiration you can just push your way through.

With regard to where you went wrong in your original posting, it is in the second line. You have writen a limit of the form $\infty -\infty$ which is indeterminate. Remember, $x$ is fixed, so $u(x), v(x)$ are also fixed, therefore, the factor $\frac{u(x+\Delta x)}{\Delta x}$ of the first term in line 4 and $\frac{u(x)}{\Delta x}$ of the second term of line 4 have infinite limits as $\Delta x \rightarrow 0$. But really, you already had this error present from line 2.

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I want to mention another way to 'arrive' at the product rule using logarithms. It is important to recognize that this is not a proof, except when the functions $u,v$ are strictly positive. However, it is a convenient symbolic way to recall the product rule (a useful mnemonic). If $y=uv$ then

\begin{align} \log{y}&=\log{uv}&\\ &=\log{u} +\log{v}\end{align}

Hence,

$\frac{y'}{y}=\frac{u'}{u}+\frac{v'}{v}$

Now, multiply by $y$ and you get $y'=u'v+v'u$. Again, this is not a proof of the general product rule, but other posters have already given satisfactory explanations using the difference quotient definition.

I personally like this method because it also works to produce a formula for the derivative of a product of an arbitrary finite number of functions (provided that all of these functions are strictly positive!).

If $y=\prod_{j=1}^n f_j$ then

\begin{align} \log{y}&=\log{\prod_{j=1}^n f_j}&\\ &= \sum_{j=1}^n \log{f_j}\end{align}

therefore,

\begin{align} \frac{y'}{y}= \sum_{j=1}^n \frac{f_j'}{f_j} \end{align}

hence

\begin{align} y'&=\sum_{j=1}^n \frac{yf_j'}{f_j}\\ &= \sum_{j=1}^n \left(f_j'(x)\prod_{\stackrel{1\leq i\leq n}{i\neq j}}f_j(x)\right) \end{align}