Yes, this follows from classification of Seifert fibered spaces. In fact, you can change the hypothesis to be "the fiber is torsion in $\pi_1(X)$" and still get the result.  $\newcommand{\RR}{\mathbb{R}}$

Suppose that $X$ is reducible. Thus $X$ has $S^2 \times \RR$ geometry and the fiber is not torsion.

Suppose that $X$ has boundary. The boundary components are tori. By "one-half lives, one-half dies" there is a unique slope $\gamma$ that dies in homology. The hypothesis implies that $\gamma$ is the Seifert slope. The disk theorem tells us that $\gamma$ bounds a disk. Irreducibility tells us that $X$ is a solid torus. We appeal to the classification of fiberings of the solid torus and find that the fiber is not torsion.

Suppose that $X$ is algebraically toroidal. If the immersed torus is a union of fibers then the fibers are not torsion. If the torus is transverse to the fibers then it is embedded. In this case $X$ has $\mathbb{E}^3$ geometry, and the fibers are not torsion.

So, $X$ is Seifert fibered, closed, irreducible, and (algebraically) atoroidal. Thus it has $S^3$ geometry.

Yes, this follows from classification of Seifert fibered spaces. In fact, you can change the hypothesis to "the fiber is torsion in $\pi_1(X)$" and still get the result. $\newcommand{\RR}{\mathbb{R}}$

Suppose that $X$ is reducible. Thus $X$ has $S^2 \times \RR$ geometry and the fiber is not torsion.

Suppose that $X$ has boundary. The boundary components are tori. By "one-half lives, one-half dies" there is a unique slope $\gamma$ that dies in homology. If the fiber is torsion, a power dies in homology, so in fact $\gamma$ is the fiber slope. The disk theorem tells us that $\gamma$ bounds a disk. Irreducibility tells us that $X$ is a solid torus. We appeal to the classification of fiberings of the solid torus and find that the fiber is not torsion, a contradiction.

Suppose that $X$ is algebraically toroidal. If the immersed torus is a union of fibers then the fibers are not torsion. If the torus is transverse to the fibers then it is embedded. In this case $X$ has $\mathbb{E}^3$ geometry, and the fibers are not torsion.

So, $X$ is Seifert fibered, closed, irreducible, and (algebraically) atoroidal. Thus it has $S^3$ geometry.

Yes, this follows from classification of Seifert fibered spaces.  $\newcommand{\RR}{\mathbb{R}}$

Suppose that $X$ is reducible. Thus $X$ has $S^2 \times \RR$ geometry and the fiber is essential. This contradicts the hypothesis.

Suppose that $X$ has boundary. The boundary components are tori. By "one-half lives, one-half dies" there is a unique slope $\gamma$ that dies in homology. The hypothesis implies that $\gamma$ is the Seifert slope. The disk theorem tells us that $\gamma$ bounds a disk. Irreducibility tells us that $X$ is a solid torus. This is a contradicts the hypothesis.

Suppose that $X$ is algebraically toroidal. If the immersed torus is a union of fibers then the fibers are essential, a contradiction. If the torus is transverse to the fibers then it is embedded. In this case $X$ has $\mathbb{E}^3$ geometry, a contradiction.

So, $X$ is Seifert fibered, closed, irreducible, and (algebraically) atoroidal. Thus it has $S^3$ geometry.

Yes, this follows from classification of Seifert fibered spaces. In fact, you can change the hypothesis to be "the fiber is torsion in $\pi_1(X)$" and still get the result. $\newcommand{\RR}{\mathbb{R}}$

Suppose that $X$ is reducible. Thus $X$ has $S^2 \times \RR$ geometry and the fiber is not torsion.

Suppose that $X$ has boundary. The boundary components are tori. By "one-half lives, one-half dies" there is a unique slope $\gamma$ that dies in homology. The hypothesis implies that $\gamma$ is the Seifert slope. The disk theorem tells us that $\gamma$ bounds a disk. Irreducibility tells us that $X$ is a solid torus. We appeal to the classification of fiberings of the solid torus and find that the fiber is not torsion.

Suppose that $X$ is algebraically toroidal. If the immersed torus is a union of fibers then the fibers are not torsion. If the torus is transverse to the fibers then it is embedded. In this case $X$ has $\mathbb{E}^3$ geometry, and the fibers are not torsion.

So, $X$ is Seifert fibered, closed, irreducible, and (algebraically) atoroidal. Thus it has $S^3$ geometry.