So the short answer is that there is not such a model structure. The difficulty arises in trying to show that the class of weak equivalences has all of the necessary properties; in particular, even two-of-three does not hold for the naive definition. The first difficulty arises even before that: on ordinary simplicial sets we can arrange for a model of every set that is "minimal" on the $\pi_0$-level, meaning that every connected component has exactly one $0$-simplex. In simplicial commutative monoids we can no longer do this. However, we could assume that in order to be a weak equivalence we need to be a $\pi_*$-isomorphism when choosing any (coherently chosen) basepoints.
For the purposes of our discussion we are going to assume that $\pi_*$-equivalences use the model $S^n = \Delta^n/\partial\Delta^n$. (This is the model that most closely mimicks the boundary maps in the Dold-Kan correspondence.) Now let $X = S^2$, and let $Y$ be $S^2$ with an extra $0$-simplex connected by a $1$-simplex to the original basepoint. (So it looks like a balloon on a string.) We define a map $X\rightarrow Y$ to be the inclusion of $S^2$ in the obvious manner, and a map $Y\rightarrow X$ to be collapsing the extra $1$-simplex back down. Then the composition of these two maps is the identity on $X$, so obviously a weak equivalence. The map $X\rightarrow Y$ is also a weak equivalence, because adding the "string" can't add any new homotopy groups to $X$. However, the map $Y\rightarrow X$ is not a weak equivalence, as $\pi_2Y$ based at the extra point is a one-point set but $\pi_2X$ at its image is a two-point set.
The problem arose because in order to show that $\pi_*$ was invariant of basepoint in the usual Kan complex model we needed to be able to "pull back" simplices along paths in the simplicial set, which used the Kan condition. The new model does not have such a condition, and thus we can't necessarily pull things back.
Another observation along these lines. Take any connected simplicial set $X$, and let $Y$ be $X$ with a "string" added to it at any basepoint. Then $*\rightarrow Y$ (including into the new point) is a weak equivalence, and $X\rightarrow Y$ (including into itself) is a weak equivalence. Thus in the homotopy category, $X$ is isomorphic to a point (and thus the homotopy category is just the category of sets) ... which is presumably not desired.
-- The Bourbon seminar
I'm not aware that the model category you want has been constructed. But it seems like an interesting question. You should ask Julie Bergner if she has thought of anything along these lines.
I don't have an answer, but I will think out loud for a bit.
(I'll be a little vague by what I mean by "space", but I probably mean "simplicial sets" here.)
I will assume that your property 3 should say: "The weak equivalences between fibrant objects (i.e., between Segal spaces) are the DK-equivalences." I would also like to throw in an additional property:
4. The trivial fibrations between Segal spaces are maps $f:X\to Y$ which are DK-equivalences, Reedy fibrations, and such that the induced map $f_0:X_0\to Y_0$ on $0$-spaces is surjective. (Note: since $f$ is a Reedy fibration, $f_0$ is a fibration of spaces.)
Yes, this comes out of thin air ... but it's modelled on the trivial fibrations in the "folk model structure" on Cat. You could go further, and posit that fibrations between Segal spaces are Reedy fibrations such that $f_0$ is surjective.
Given a space $U$, let $cU$ denote the "$0$-coskeleton" simplical space, with $(cU)_n=U^{\times (n+1)}$. If $U$ is a fibrant space, then $cU$ is Reedy fibrant; if $U\to V$ is a fibration, $cU\to cV$ is a Reedy fibration. Furthermore, $cU$ clearly satisfies the Segal condition.
Thus, if $g:U\to V$ is a surjective fibration of spaces, $cg: cU\to cV$ should be a trivial fibration in our model category, according to 4.
The functor $c$ is right adjoint to $X\mapsto X_0$: that is, maps of simplicial spaces $X\to cU$ are naturally the same as maps $X_0\to U$ of spaces.
Putting all this together, we discover that, if such a model category exists, a cofibration $f: A\to B$ should have the following properties: the map $f_0 : A_0\to B_0$ is a cofibration of spaces, and $B_0=B_0'\amalg B_0''$ so that $f_0$ restricts to a weak equivalence $A_0\to B_0'$, and such that $B_0''$ is homotopy discrete (i.e., has the weak homotopy type of a discrete space).
In particular, a necessary condition for $B$ to be cofibrant is that $B_0$ is homotopy discrete.
This is a pretty restrictive condition on cofibrations, but it does not seem impossible. If there actually was a model category with all these properties, it appears that the class of fibrant-and-cofibrant objects would be what you might call the quasi Segal categories. These are the Segal spaces $X$ such that $X_0$ is homotopy discrete. Cofibrant replacement of a Segal category would give a DK-equivalent quasi-Segal category.
That would be a pleasing outcome, and probably along the lines of what you're looking for.
Best Answer
Here is a proof that any autoequivalence of (the $\infty$-category of) filtered spectra is naturally equivalent to a suspension functor. Since filtered spectra are equivalent to simplicial spectra, this translates back. The argument takes some work but the following sketches it.
Autoequivalences preserve categorical properties. Therefore, they preserve homotopy limits and colimits, hence the zero object, cofibers, sums, and so on.
In particular, autoequivalences preserve compact objects. A filtered spectrum $X_0 \to X_1 \to \dots$ is compact if and only if each $X_i$ is compact and the maps $X_n \to X_{n+1}$ are equivalences for $n >> 0$. This condition is sufficient because such objects are built in finitely many stages from free objects $0 \to \dots \to 0 \to S^n \to S^n \to \dots$, and you can check necessity by mapping out to diagrams of objects of the form $Y \to Y \to \dots \to Y \to 0 \to 0 \to \dots$.
Because filtered spectra are stable, they get mapping spectra as part of the categorical structure. If $X$ and $Y$ are compact then $map(X,Y)$ is a finite spectrum.
Similarly because filtered spectra are a cocomplete, stable $\infty$-category, they also get a left-tensoring over spectra: $Z \otimes (X_n) \simeq (Z \otimes X_n)$. Any autoequivalence must preserve this left-tensoring.
For any filtered spectrum $X$, we get a ring spectrum $end(X)$, and if $X$ is compact then by the above we find that for any ring spectrum $R$ we can calculate $end(X) \otimes R \simeq end_R(R \otimes X)$.
Let's define the indicator $g(n,m)$ by $$ g(n,m)_k \simeq \begin{cases} \Bbb S &\text{if }n \leq k \leq m,\\0 &\text{otherwise,}\end{cases} $$ with all transition maps being identities. Note that $end(g(n,m)) \simeq \Bbb S$. (I'll allow $m=\infty$ in this definition.)
For any field $k$, any filtered $Hk$-module $Y$ is a direct sum of shifts of indicators: $\Sigma^r Hk \otimes g(n,m)$. As a result, $end_{Hk}(Y) \simeq Hk$ if and only if if $Y$ has one such summand: it is a shift of an indicator.
Here is a lemma: a compact filtered spectrum $X$ satisfies $end(X) \simeq \Bbb S$ if and only if it is equivalent to a suspension $\Sigma^r g(n,m)$ for some $r, n, m$. To check this, we note that for any field $k$, the filtered $Hk$-module $Hk \otimes X$ satisfies $end_{Hk}(Hk \otimes X) \simeq Hk$ by point 6, and hence it is an indicator by point 7. For this to be true, the universal coefficient theorem shows that $H\Bbb Z \otimes X$ must a shifted indicator $\Sigma^r H\Bbb Z \otimes g(n,m)$, and the Hurewicz / Whitehead theorems then let us construct an equivalence $g(n,m) \to X$.
As a result, the family $\{\Sigma^r g(n,m)\}$ of (shifted) indicators is preserved (up to equivalence) by any autoequivalence.
Now we note that $$ map(\Sigma^r g(n,m), \Sigma^s g(p,q)) \simeq \begin{cases} S^{s-r} &\text{if }p \leq n \leq q \leq m,\\ S^{s-r-1} &\text{if }n < p \leq m + 1, $m < q,\\ 0 &\text{otherwise.} \end{cases} $$
I claim that property 10 allows us to identify $\Sigma^r g(0,\infty)$ among indicators, and therefore any autoequivalence must take $g(0,\infty)$ to something equivalent to $\Sigma^r g(0,\infty)$ for some $r$. In previous edits I've taken shortcuts with mixed success; here is a lengthier argument.
Similarly, for any other indicator $h$, $map(h, \Sigma^r g(0,\infty)) \simeq \Bbb S$ if and only if $h \simeq \Sigma^r g(n,\infty)$ for some $n$.
If $h_1 \simeq \Sigma^r g(n_1,\infty)$ and $h_2 \simeq \Sigma^r g(n_2,\infty)$ are two such indicators as in point 11, we have that $n_1 \geq n_2$ if and only if $map(h_1,h_2) \simeq \Bbb S$.
This lets us show that, up to equivalence, any autoequivalence of this category must, up to equivalence, take the tower $$ g(0,\infty) \leftarrow g(1,\infty) \leftarrow g(2,\infty) \leftarrow g(3,\infty) \leftarrow \dots $$ to a tower equivalent to $$ \Sigma^r g(0,\infty) \leftarrow \Sigma^r g(1,\infty) \leftarrow \Sigma^r g(2,\infty) \leftarrow \Sigma^r g(3,\infty) \leftarrow \dots $$
This shows that, up to equivalence, the autoequivalence is equivalent to the shift $\Sigma^r$ on the subcategory spanned by $g(n,\infty)$. However, this is a set of generators for the category of filtered spectra: every other filtered spectrum is built from them by hocolims. Therefore, the autoequivalence must be equivalent to the shift $\Sigma^r$.
As a final note, there is an equivalence of categories between $Fun^L(Fil(Sp), Fil(Sp))$ and the functor category $Fun(\Bbb N \times \Bbb N^{op}, Sp)$ by a compact-generators argument. Therefore, the space of autoequivalences can be determined by looking at the space of self-maps of the object representing the identity functor in this category.