You're quite right that "the $\mu$-operator breaks $\Delta^0_1$-ness." As an explicit example, the formula $\theta(x,y)\equiv$ "the $x$th Turing machine with input $x$ halts within $y$ steps" is $\Delta^0_1$, but the unary relation $$H(x)\equiv \mbox{"}0\not=\mu y.\theta(x,y)\mbox{"}$$ is not $\Delta^0_1$.
However, this doesn't contradict Post's theorem. The mistake is the following:
I always thought “general recursive/computable” meant the total functions closed under the primitive recursive functions and minimization, not the (non total) partial functions.
That's not correct - (general) recursive functions are allowed to be partial. In particular, there is a recursive enumeration of the recursive functions, but there is no such enumeration of the total recursive functions.
Indeed, this is sort of the driving observation behind recursion theory. Remember that diagonalization tells us that we can never have an "effective" notion of "total effective function" - otherwise, the function $f$ given by $f(x) = 1+ t_x(x)$, where $t_x$ is the $x$th effective function according to this notion, would be simultaneously total effective and not total effective. Whenever we have a "concretely defined" class of total recursive functions, it falls short of being the whole class of total recursive functions.
By contrast, once we allow partiality this issue goes away. We can whip up a recursive enumeration $(\varphi_e)_{e\in\mathbb{N}}$ of the (partial) recursive functions (essentially, this is a universal Turing machine), and the function $$\theta(x)=1+\varphi_x(x)$$ is indeed recursive, but there's no contradiction: we have $\theta=\varphi_e$ for some $e$ such that $\varphi_e(e)$ is undefined. Our diagonalizing function $\theta$ is also therefore undefined at $e$, and so there's no issue. In fact, if we spend enough time thinking about this situation (and get a bit clever), we'll eventually arrive at the recursion theorem.
Of course, ease of use isn't a fully satisfying motivator for a shift in the notion of "fundamental object." But recursion theory is supposed to model algorithms, and the simple fact is that there are algorithmic procedures which don't halt, and - while we can easily check that a purported algorithm is "grammatically correct" - there is no good way to determine if an algorithm is going to halt on a given input. So in fact partial functions correspond better to what we care about than total functions.
Since you mention graphs: a (unary, for simplicity) recursive function can be identified with its graph $G_f:=\{(x,y): f(x)=y\}$, and a relation $R\subseteq\mathbb{N}\times\mathbb{N}$ is the graph of some recursive function iff:
If the function is total then its graph is recursive, but otherwise it's merely r.e. - this is a consequence of the second clause, since it means that we can compute $f(x)$ by searching for the unique $y$ such that $(x,y)\in G_f$. By contrast, if we drop the second clause this breaks: a total r.e. relation need not be recursive.
Best Answer
If a formula $\phi(x_1, ... x_n)$ is in both $\Sigma_1, \Pi_1$, then one can define a Turing machine to determine whether it is true or false. Namely, in parallel, search for a collection of parameters that makes true the existential formula, and search for a collection of formulas that makes false the universal formula. If the first happens, return true; if the second happens, return false. One of these must exist, so the Turing machine always halts.
(The set of $x_1,...x_n$ such that $\phi(x_1, ... , x_n)$ is valid if $\phi$ belongs to $\Sigma_1$ is, by contrast, is only recursively enumerable.)
By contrast, since the action of any Turing machine is simulable by existential formulas in first-order logic (i.e. there exists a number $k$ such that $M$ halts in $k$ steps), any language which is recursively enumerable can be expressed by existential formulas. Any language whose complement is recursively enumerable can similarly be expressed by universal formulas (by the analog of deMorgan's laws). So any recursive language (i.e., one which is both recursively enumerable and whose complement is r.e.), can be expressed in both ways.