There is some truth in what you are implying, but you need to sort things out.
Let a boolean expression be a combination of binary variables, the operators AND $\land$ , OR $\lor$ and NOT $\neg$ and the constants 0 and 1.
Then, every computable function can be expressed as a boolean expression.
$SAT$ is defined as the set of satisfiable boolean expressions. The $\mathbb{NP}$-complete problem is to decide if a boolean formula, that is given as input, is satisfiable or not.
Now, let's combine the two. Every problem can be decided by a boolean formula that determines membership in the problem's language. The SAT problem is $\mathbb{NP}$-complete, therefore, every problem is $\mathbb{NP}$-complete.
The above proof is of course false, since no single complexity class, that sets bounds on resources, can contain all problems. So, where's the catch?
Suppose that you have an instance $a$ of a problem $A$. In order to convert it to a boolean formula, i.e. reduce it to $SAT$, a process must be followed. This process requires additional time, therefore this process combined with an algorithm for SAT does not yield a polynomial time non-deterministic algorithm for every problem, but only for problems in $\mathbb{NP}$.
Suppose that we are not interested in time, but in computability. So, we don't care how long the algorithm takes to convert any instance of any problem to a boolean formula, but if it can do so in finite time. Since we know that this is possible, a computation model using only boolean formulas is Turing equivalent and can compute any function computable by a Turing machine.