Let $f: S\to S$ be a homeomorphism of a compact connected surface (possibly with boundary).
Theorem.
If $\chi(S)<0$ then the mapping torus $M=M_f$ of $f$ is a Seifert manifold if and only if the mapping class of $f$ is periodic, i.e., $f$ is isotopic to a periodic homeomorphism.
If $\chi(S)\ge 0$ then $M$ is Seifert unless $S$ is the torus and eigenvalues of the action of $f$ on $H_1(S, {\mathbb R})$ are not roots of unity.
Proof. I will prove only Part 1 since the proof is already way too long. Recall that the surface $S$ admits Thurston-Nielsen decomposition with respect to $f$, i.e., a finite collection (possibly empty) of simple loops $L_i\subset S$ which are pairwise disjoint, pairwise non-isotopic and not isotopic to the boundary of $S$ (if there is any), such that:
a. $f$ preserves the multiloop $L=\cup_i L_i$. In particular, there exists $N$ such that $f^N$ preserves
each loop $L_i$ and each component $S_j$ of $S\setminus L$.
b. The homeomorphism $f_j^N= f^N|S_j$ is homotopic to a homeomorphism $g_j: S_j\to S_j$ which is either periodic (by taking larger $N$ we can assume that such $g_j$ is the identity) or is pseudo-Anosov.
A homeomorphism $f$ is called reducible if $L$ is nonempty.
A very nice proof of this fundamental theorem can be found for instance in the book by Casson and Bleiler "Homeomorphisms of surfaces after Nielsen and Thurston."
Since a manifold is Seifert if and only if it has finite cover which is Seifert, we can assume that $S$ is oriented and $f=f^N$ (replacing $f$ with $f^N$ amounts to passing to a finite cover of $M$).
We now define tori $T_i$ which are mapping tori of the restrictions $f|L_i$. Then, by construction, the manifold $M=M_f$ is the union of submanifolds with boundary $M_j=M_{f|S_j}$, which meet along the tori $T_i$. As it was observed in the other answer, if $f|S_j$ is pseudo-Anosov, then $M_j$ is hyperbolic and, hence, $M$ cannot be a Seifert manifold in this case. (For instance, this can be seen from the fact that $\pi_1(M_j)$ has trivial center, while $\pi_1$ of a Seifert manifold always has nontrivial center, after passing to a finite cover, unless this cover is the 3-sphere.)
It remains to analyze the case when each $f|S_j$ is isotopic to the identity. Let $A_i=\eta(L_i)$ denote a small annular neighborhood of $L_i$ in $S$. Let $S_j'$ denote the complement in $S_j$ to all the annuli $A_j$ which it meets. By isotopying $f$ further, we can assume that $f_j=f|S_j'$ is the identity while the restriction $f|A_i$ is an iterated Dehn twist $D^{n_i}$ along the loop $L_i$. Then the mapping torus of $f_j$ is the product
$S_j'\times S^1$. If some $n_i$ equals zero, then $f|A_i=Id$ and we can eliminate the loop $L_i$ from $L$ without changing topology of $M$ and periodicity (or lack of thereof) of $f$.
Each annulus $A_i$ has two boundary circles which I will denote $A_i^+, A_i^-$; accordingly, each manifold $M_{f|A_i}\cong A_i\times S^1$ has two boundary tori $T_{i}^+, T_i^-$ which are the mapping tori of $f|A_i^\pm$. Each loop $A_i^\pm$ is a boundary loop of a unique component of
$$
\bigcup_j S_j'
$$
which I will denote $S_i^+$ and $S_i^-$ accordingly. (Note that it might happen that $S_i^+=S_i^-$, for instance, if $L$ is a single loop which does not separate $S$.) Accordingly, the (mapping) torus $T_{i}^\pm=M_{f|A_i^\pm}$ is a boundary component of the product manifold $M_i^\pm=M_{f|S_i^\pm}$, the mapping torus of the homeomorphism $f$ restricted to $S_i^+$ or $S_i^-$. The manifold $M_i\pm$ has canonical product structure (since $f$ restricts to the identity on $S_i^\pm$). Therefore, for each torus $T_{i}^+$ we obtain a canonical system of generators of the homology group: "Horizontal'' loop $a_{i}^+$ corresponding to the loop $A_i^+\subset S$ and the "vertical'' loop $b_i^+$ corresponding to the $S^1$-factor of the decomposition $M_i^+=S_i^+\times S^1$.
The same applies to the torus $T_{i}^-$, where we switch all pluses to minuses.
Now, consider these loops $a_i^\pm, b_i^\pm$ in the product $M_{f|A_i}=T^2\times [0,1]$. The loops $a_i^+, a_i^-$ are, of course, isotopic to each other in this product manifold, since they correspond to isotopic curves $A_i^+, A_i^-$ on the surface $S$. I will denote by $[a_i]$ their common homology class in this product manifold.
However, this is not the case for for the loops $b^+_i, b^-_i$: From the definition of the Dehn twist we obtain that
$$
[b_i^-]= [b_i^+]\pm n_i [a_i]
$$
the plus or minus depend on which boundary circle of $A_i$ we marked with $+$ and which with $-$; the number $n_i$ here is the power of the Dehn twist that we use.
The bottom line is that when under the gluing $M_i^+$ and $M_i^-$ along $A_i\times S^1$, the fibers of the (Seifert) fibration of $M_i^+$ do not match (up to isotopy) fibers of the Seifert fibration of $M_i^-$.
It is important to note here that the manifolds $M_i^\pm$ here admit unique, up to isotopy, Seifert fibrations, namely, the ones coming from their product decompositions $S_i^\pm \times S^1$, since each component of the surface $S\setminus L$ has negative Euler characteristic. (Here we are using the fact that $S$ is not the torus: cutting torus along a loop results in the annulus $A$ and the product $A\times S^1$ admits infinitely many, up to isotopy, circle fibrations.) This uniqueness theorem should be in Hempel's book on 3-manifolds and in Orlik's book on Seifert manifolds. In particular, up to isotopy, we can talk about the Seifert fibration of the manifold $M_i^\pm$.
The product regions $A_i\times S^1$ of course, admit infinitely many Seifert fibrations; to remedy this, we adjoin each $A_i\times S^1$ to the product manifold $S_i^+\times S^1$. As the result, $M$ is obtained by gluing product manifolds $M_j\cong S_j\times S^1$ along their boundary tori in such a fashion that all the gluing maps do not preserve the (up to isotopy) fibers of the circle fibrations of the manifolds $M_j$.
Now, we can finish the proof of Part 1 of the theorem with the following lemma which, I remember seeing in the book by Jaco and Shalen "Seifert fibered spaces in 3-manifolds", Memoirs of Amer. Math. Soc. 220 (1979).
Lemma. Suppose that $M$ is a 3-dimensional manifold obtained by gluing oriented Seifert manifolds $M_j$ along their incompressible boundary tori, where each $M_j$ admits a unique Seifert fibration. Then $M$ is Seifert if and only if all gluing maps preserve (up to isotopy) fibers of the Seifert fibrations.
Proof. I will skip the proof of one direction of this lemma since it is not needed and assume that one of the gluing maps does not preserve circle fibers. Let $T\subset M$ denote the incompressible torus corresponding to this gluing. Suppose, that $M$ is Seifert fibered; by looking at its fundamental group, it is clear that the base of this fibration has to be of hyperbolic type (i.e., it is a hyperbolic orbifold). It is a standard fact of the theory of Seifert manifolds (I am sure, it is in Jaco and Shalen) that every incompressible torus in such a fibration is "vertical'', i.e., isotopic to a torus foliated by Seifert fibers. To see the algebraic side of this statement, consider the short exact sequence of fundamental groups induced by Seifert fibration:
$$
1\to {\mathbb Z}\to \pi_1(M)\to B=\pi_1(O)\to 1,
$$
where $O$ is the base-orbifold of the fibration and the fundamental group of $O$ is understood in the orbifold sense. Since the torus $T$ is incompressible, its fundamental group yields a subgroup $Z^2$ of $\pi_1(M)$. Projection of this group to $B$ has to be abelian, and, by hyperbolicity assumption on $O$, infinite cyclic. Therefore, the intersection of $Z^2$ with the normal subgroup ${\mathbb Z}$ of $\pi_1(M)$ is a free (abelian) factor of $Z^2$. In particular, $T$ admits a foliation by circles where each fiber is homotopic to the generic fiber of the Seifert fibration of $M$. By working more, one promotes this to an isotopy of $T^2$.
Instead of isotopying the torus we can isotope Seifert fibration itself. Therefore, splitting $M$ along $T$ results in one or two Seifert manifolds and the gluing map preserves Seifert fibrations. Continuing inductively with respect to all the boundary tori of the manifolds $M_i$, we obtain that each $M_i$ admits a Seifert fibration and gluing maps preserve these fibrations. Since Seifert fibration of each $M_i$ was unique (up to isotopy), we are done. QED
Best Answer
The key reference here is Jean Cerf, "Groupes d’automorphismes et groupes de difféomorphismes des variétés compactes de dimension 3", Bull. Soc. Math. France (1959). The full text is available here.
Let $M$ be a closed 3-manifold, $G$ its group of self-homeomorphisms, $H$ its group of self-diffeomorphisms. (All orientation preserving for convenience.) He proves that $\pi_n(G,H) = 0$ for all $n \geq 0$, thus that the inclusion $H \hookrightarrow G$ is a weak homotopy equivalence - assuming Smale's conjecture (now theorem, due to Hatcher) that the inclusion $SO(4) \hookrightarrow \text{Diff}^+(S^3)$ is a homotopy equivalence. (In fact, he shows that the inclusion is a homotopy equivalents, but the arguments are a bit more delicate.)
For a manifold $M$ with boundary, $H$ and $G$ as above should be the automorphisms restricting to the identity on the boundary. The idea is to show that, if you have a decomposition $M = M_1 \cup M_2$, with $M_1 \cap M_2$ a properly embedded surface, having $H_i$ be $(n-1)$-connected in $G_i$ (for both $i$) is equivalent to having $H$ be $(n-1)$-connected in $G$.
Now we want to induct. Say that a manifold homeomorphic to $D^3$ is order 0; and a manifold that decomposes into order $(n-1)$-pieces is order $n$ if we can't decompose it into pieces smaller than order $(n-1)$. But by Smale's theorem, it's true that $\pi_i(G,H) = 0$ for all $i \geq 0$ when $M = D^3$. Because every 3-manifold has finite order (look at Heegaard decompositions), this proves the theorem.
As for why the third paragraph is true, this is Cerf's "Lemma 0", at the very end of his paper.