Can we say that $\frac{0}{0}$ is every number? Suppose we have an equation $ab=0$. This equation is true when statements $a=0$ or $b=0$ are true.
If $a=0$, then $b=\frac{0}{0}$. That means $b$ could be any number for $ab=0$ to be true. If the set which groups all the numbers is the complex set, then $b$ will every number within $\mathbb{C}$, so $\forall z\in\mathbb{C}:b=\frac{0}{0}=z$. Therefore, $\frac{0}{0}$ is every number.
I know it really is not defined as number but conceptually it is every number, right?
Is this right, or am I missing something?
 A: No, it is not every number.
You are correct that for every number $x$ we have that $0 \cdot x = 0$, but we want the ratio to be a single number, because we want it to act like a function ... so it is exactly because we don't have a unique solution to $0 \cdot x = 0$  that we say that $\frac{0}{0}$ is not a number ... and certainly not all numbers.
Instead, we say that $\frac{0}{0}$ is 'indeterminate'.  Interestingly, if we define the relationship over natural numbers '$a$ divides $b$' as there exists some natural number $c$ such that $a \cdot c = b$, then that means that we can also say that '0 divides 0'.  But that still doesn't mean that $\frac{0}{0}$ is defined!
A: There are some contexts where it is helpful to consider multiple-valued fractions $\,x = a/b\,$ that denote the complete solution set of $\,b\,x = a.\,$ Then  $0/0$ denotes the solution set of $\,0x = 0,\,$ which is satisfied by every element of the domain. 
One simple context where they arise is modular arithmetic, e.g. consider the fraction $\,x\equiv 6/2\pmod{\!10}.\,$  We have $\,2x\equiv 6\pmod{\!10}$ $\iff 2x = 6+10k\,$ $\iff x = 3 + 5k$ $\iff x\equiv 3\pmod{\!5}$ $\iff x\equiv 3,8\pmod{\!10}.\, $ So this fraction takes two values mod $10.\,$ Such fractions arise naturally in the fractional form of the extended Euclidean GCD algorithm . 
For example, see the Remark in the linked post, where we bootstrap the Euclidean modular reduction process with  $\, 0/18 \equiv 0/0 \pmod{\!18}.\ $ We could avoid using $\,0/0\,$ there, but since multiple-valued fractions occur throughout the algorithm, it is natural to allow all such fractions.
That said, generally such notions are avoided in elementary contexts since it is very easy to make mistakes if one does not have a solid grasp on the fundamentals.
A: Short answer:  No.
Slightly longer answer:  Still no, but here's a slightly related concept which I think is "as close as we can get" to an affirmative answer to your question, and maybe you already know this and it's why you asked the question in the first place.  But anyway..
It is possible to get $0/0$ from direct substitution in a limit whose value can be any (fixed) number.
A well-known limit typically introduced no later than a first-semester calculus course is:
$$ \lim_{x\to 0} \frac{\sin x}x = 1$$
Direct substitution of $x=0$ into this limit gives us $\dfrac{\sin 0}0 = \dfrac00$, which is indeterminate.  But the actual value of the limit is $1$.  Therefore we can multiply $\dfrac{\sin x}x$ by any real (or complex) constant $c$ to get $c$ as the limiting value as $x \to 0$, while direct substitution still gives $0/0$.
$$ \lim_{x\to0} \frac{c \sin x}x = c \lim_{x\to0} \frac{\sin x}x = c \cdot 1 = c,$$
and direct substitution gives us the indeterminate form $\dfrac{c \cdot 0}0 = \dfrac00$.
A: When we say that $ab=0$ we define $a$ and $b$ as being unique numbers. So when saying $b$ is all the numbers we are saying that $b$ is not unique. That leads to a contradiction so $b$ is undefined.
A: Ultimately, it's not a number, or every number, because it isn't useful to interpret it that way.  It might be interesting from a philosophical perspective, but since we generally can't move it into another context and expect it to do anything predictable, we typically treat it as undefined.  (See Bill Dubuque's answer for an interesting counterpoint, however.)
Often, the expression $\frac{0}{0}$ might show up as a result of comparing two continuously varying values that just happen to equal $0$ simultaneously—something like
$$
\frac{f(a)}{g(a)}
$$
if both $f(x)$ and $g(x)$ happen to have zeros at $x = a$.  But for such situations, we use limits, and instead of the foregoing, we instead substitute
$$
\lim_{x \to a}\frac{f(x)}{g(x)}
$$
In some cases, we have $\infty$ in place of $a$, to refer to the ultimate ratio of the fraction $\frac{f(x)}{g(x)}$ as $x$ increases without bound.
In the case of your expression, you have $b = \frac{0}{a}$, which is ordinarily just $0$.  But because $a$ itself is also equal to $0$, you have concluded $b = \frac{0}{0}$, which you choose to interpret as $b$ being able to take on any value.  However, you might have just concluded that $b$ can take on any value without going through the intermediary expression $\frac{0}{0}$.  And because that expression is essentially useless when performing analysis, that is exactly what mathematicians typically do: simply say that $b$ can take on any value.
A: You can say $\frac 00$ is a pink elephant that eats dinosaurs if you want.  But if you say that then you can't say $\frac 00$ is a distinct number that when multiplied by $0$ results is zero.  
So if you $\frac 00 $ is every number, you can not say $\frac 00$ is a distinct number that when multiplied by $0$ results in $0$.  "Every number" is not a distinct number.  "Every number" is a set of numbers.  "Every number" is a the set of all numbers. 
$\frac ab$, by definition, is a distinct number, $x$ so that $x*b = a$.  It is not, by definition, a set of multiple numbers.
So what we say instead is: As there are multiple numbers, $x$ such that $x*0 = 0$ and indeed for any number, $x$, we have $x*0 = 0$, there is no distinct number so that $x*0 = 0$, therefore $\frac 00$ can not be defined with the usual definition.  However we can say $X = \{x \in R| x*0 = 0\} = \mathbb R$.  But that is an entirely different statement than: $\frac 00 = \mathbb R$.
Actually if you want to get ... obtuse.  You can define $X_{a,b} = \{x \in \mathbb R| x*b = a\}$ and define $\frac ab$ as the single element of $X_{a,b}$ if $X_{a,b}$ has only a single element.  We can show that if $b \ne 0$ then any set $X_{a,b}$ will have a single element and thus $\frac ab$ is uniquely defined.  If $b = 0$ and $a \ne 0$ then $X_{a,b} = \emptyset$ and if $b=0; a=0$ then $X_{a,b} = \mathbb R$, and in either case $\frac ab$ is defined if and only if $b \ne 0$.
But that'd be silly.
====
Note:  I did say with the "usual definition".  See Bill Dubuque's answer for alternate definitions.  As well, in the extended number line, $\frac 10 = \infty$ is acceptable but then $\frac ab$ no longer is defined as the number $x$ so that $b*x =a$.  
A: You may call $\frac00$ "every number" if you like but that leads you nowhere. Because you can't use this "every number" in further arithmetic operations.
For instance, zero times "every number" is still undefined, and it would be a wrong idea to define it as zero.
A: Lets check
$\frac{0}{0}=5$
We multiply both sides of equation by $5$
$5 \cdot \frac{0}{0} = 5 \cdot 5$
$5 \cdot 0 + 0 = 25$
$0 + 0 = 25$
$0 = 25$
No, we can't. Logically $b$ cannot be a value and an operation at the same time.
A: In the ordinary number systems division by zero is undefined. It is possible to define extended number systems including objects as $\frac{1}{0}=\infty$ and $\frac{-1}{0}=-\infty$, "infinite numbers" and $\frac{0}{0}$, "indefinite number". The rules wouldn't differ that much but the system would not gain much either. The most theorems would be possible to rehabilitate, but would mostly be more ugly. I guess. 
A: You may think of the form $0/0$ as a variable; but I think that is misleading outside the context of functions. Rather, we say it is an indeterminate, or a non-determinate, or an unknown, but certainly not some number in $\Bbb C$, or even a set of all numbers as you seem to think.
A: There's actually nothing too wrong with taking $\frac{0}{0}$ to have every real number as a value. That is, as you point out, the most logical thing to do when you consider the defining equation for division:
$$b \left(\frac{a}{b}\right) = a$$
which gives, when $a = b = 0$,
$$0 \left(\frac{0}{0}\right) = 0$$
and any substitution for the putative quantity $\frac{0}{0}$ will work, and make this equation hold good true (something that does not happen when $a \ne 0$ and $b = 0$!). Moreover, in many geometric limiting contexts, this makes a lot of sense, too: if you do this, the graph of the equation
$$y = \frac{0}{x}$$
is a cross, which is the exact limiting case of the hyperbolas
$$y = \frac{a}{x}$$
as $a \rightarrow 0$.
So why isn't it done? Simple: We like to have division be a function, and not just a one-to-many (many many!) relation. Having it take all real values at even one point would destroy that. So the standard option is the alternative: limit the domain so that $(0, 0)$ is absent therefrom. This leaves $\frac{0}{0}$ undefined.
But there is nothing inherently wrong with considering such non-function operations: the inverse trigonometric functions, the square root, etc. are all most properly thought of as one-to-many relations, and they pop up in many other areas as well.
