I suggest that you should think of $\delta$ a bit differently than you do, namely either as a distribution or as a measure. I will discuss the first point of view. The notation with integrals are handy but often lead to confusion when the "integrand" is not a real function.
So, what is a distribution? Well, you could say it is something that eats nice functions (typically belonging to some nice space, like infinitely differentiable functions with compact support, $C_0^{+\infty}(\mathbf R)$) and spits out numbers. If $\phi$ is such a function and $T$ is a distribution, then this is usually denoted by $T(\phi)$ or $\langle T,\phi\rangle$. We say that $T$ acts on the test function $\phi$. Distributions should also be linear, so $T(\phi+\psi)=T(\psi)+T(\phi)$ and $T(a\phi)=aT(\phi)$. If we want to use fancy names, we could say that a distribution $T$ is a linear functional on $C_0^{+\infty}(\mathbf R)$.
Main example of distribution
Let $f$ be a function such that the integral
$$
\int_{-\infty}^{+\infty} f(x)\phi(x)\,dx
$$
makes sense for all $\phi\in C_0^{+\infty}(\mathbf R)$ (such a function is usually said to be locally integrable), then there is a natural distribution $T$ associated with $f$, namely
$$
T(\phi)=\int_{-\infty}^{+\infty}f(x)\phi(x)\,dx.
$$
Without going into details, more or less all the functions you usually meet (with not too strong singularities) can be considered as distributions in this way, and sometimes one even does not distinguish between the function $f$ and the distribution $T$. However, it is possible to define distributions that does not correspond to any function as above, so distributions are in some sense more general then functions.
Important point
A distribution $T$ is not (unlike functions) defined at specific points $x\in\mathbf R$, so that it does not make sense to talk about $T(x)$. So, how could one define something that should correspond to typically "point-wisy" operations as differentiations $f'$, a dilation $f(bx)$ ($b\neq 0$) or a shift $f(x-b)$?
Getting closer to your question
Let us concentrate on how $T(bx)$ is defined for distributions, since this is what the question is about. Before doing so, let us see what happens in the case of when $T$ is given as in the Main example above. Thus, we want to somehow relate $T$ above with
$$
\int_{-\infty}^{+\infty}f(bx)\phi(x)\,dx.
$$
The change of variables $x\mapsto u=bx$ then transforms this integral into
$$
\frac{1}{|b|}\int_{-\infty}^{+\infty}f(u)\phi(u/b)\,du.
$$
But this integral can be written as
$$
\frac{1}{|b|}T(\phi(\cdot\,/b)),
$$
i.e. the factor $1/|b|$ times the action of our original distribution $T$ acting on the dilated test function. As is common in distribution theory (this is done for derivatives of distributions, shifts, Fourier transforms, ...), we want distributions to work like functions, and make a definition:
Definition of dilation of distributions
Given a distribution $T$, we define its dilation $T_b$ as the distribution
$$
T_b(\phi)=\frac{1}{|b|}T(\phi(\cdot\,/b)).
$$
(Just a reminder: To define a distribution, we need to say how it acts on test functions.)
Over to $\delta$ and the answer to your question
The delta distribution $\delta$ is the distribution defined by
$$
\delta(\phi)=\phi(0)
$$
i.e. it acts on a test function by returning the value of the test function at zero. By definition, the dilation of the delta function (which you wrote $\delta(x/b)$, but which I will write $\delta_b$ according to the definition above)
$$
\delta_b(\phi)=\frac{1}{|b|}\delta(\phi(\cdot\,/b))=\frac{1}{|b|}\phi(0).
$$
The second line is just the definition of $\delta$. I hope this answers your question.
Final comments
1) I hope it is clear to you that there does not exist a function $f$ such that
$$
\int_{-\infty}^{+\infty}f(x)\phi(x)\,dx=\phi(0)
$$
for all $\phi\in C_0^{+\infty}(\mathbf R)$.
2) The choice of space $C_0^{+\infty}(\mathbf R)$ above could have been some other space.
It depends on your definition of Lebesgue Integrable...
If $\int \delta d \lambda := \lim_{n \to \infty} \int \delta_n d \lambda,$ then, since the limit exists and is 1, $\delta$ is integrable.
However, usually the Lebesgue integral is defined as $$\int \delta d \lambda := \sup\left\{ \sum_{i=1}^n (b_i-a_i)\alpha_i \right\},$$ where the $\sup$ runs over all functions $\varphi(t) := \sum_{i=1}^n \alpha_i \cdot 1_{(a_i,b_i)}(t),$ that satisfy $0 \leq \varphi(t) \leq \delta(t).$ (Here $1_S(t)$ is the indicator function of the set $S$.) In this definition, $\int \delta d \lambda = 0$, so $\delta$ is integrable for a stupid reason: it's 0 almost everywhere.
PS: A generally difficult problem in analysis is to identify when $\lim \int f_n = \int \lim f_n.$ With your $\delta$ and $\delta_n$ functions, you have an example of when the two sides are not equal.
Best Answer
This is incorrect. One can not work with powers of the Dirac delta function, as if it were a ordinary function. To see that such a calculation goes wrong, let us use the simplest representation of the delta function, namely a block. I.e. we define the delta function $\delta(x)$ as $1/\epsilon$ for $-\epsilon /2 < x < +\epsilon /2$ and $0$ elsewhere. We now get:
$$\int _{-\infty} ^{\infty} \delta(x)^n f(x)dx = \frac {1}{\epsilon ^n} \int_{-\epsilon/2} ^{\epsilon/2}f(x)dx = f(0) / \epsilon^{n-1}$$
As required we now take the limit of $\epsilon$ to $0$, and then we see that for all powers of $n$ larger than $1$ the result diverges, i.e. goes to $+\infty$.