I think there is a possible reduction of this question. Consider a generic smooth torus embedding $f: T^2 \to S^1\times S^4$, then the projection to $S^1$ gives a Morse function on $T^2$. Because of the ambient dimension (codimension 3), I think one may isotope $f$ to cancel critical points and obtain an isotopic embedding $f': T^2\to S^1\times S^4$ such that the projection to $S^1$ has no critical points when restricted to $T^2$ if the map $f_{|T^2}:T^2\to S^1$ is homotopically non-trivial. Thus $f'^{-1}(z)$ is a union of circles in $S^4$ for all $z\in S^1$. The homotopy type of $f'$ should be determined by the number of circles, or correspondingly the index of the image $\pi_1(T^2)\to \pi_1(S^1)$.

If the configuration space of $k$ circles in $S^4$ has $\pi_1$ isomorphic to the symmetric group $S_k$, then I think that one should be able to show that any two such maps are isotopic.

For $k=1$, ie the case that $f'^{-1}(z)$ has a single circle, I think one can show that all such tori are isotopic. Because this is codimension $3$ in $S^4$ it is unknotted.

  Let $f: S^1 \times S^1 \to S^4$ be a map so that $f(z,S^1)$ is embedded for all $z$, then we may think of $f$ as a loop in in the configuration space of embeddings of a circle in $S^4$. This loop may be achieved by a 1-parameter family of diffeomorphisms by ambient isotopy. Let $D$ be an embedded disk bounding $\gamma= f(1,S^1)$, and let $D'$ be the disk obtained after the ambient isotopy. By the 4-dimensional light-bulb theorem (Theorem 1.10), $D$ is isotopic to $D'$ rel $\gamma$. So this isotopy may be achieved by a 1-parameter isotopy of disks. But any such isotopy is homotopic to the constant isotopy. Consider a point on the disk, the point gives a loop in $S^4$ which is contractible, and hence one may assume that the isotopy fixes this point. Then one may also assume that the tangent space to the disk at the point is fixed, and then that the disks are fixed by shrinking them down to the tangent space. Hence I think that the tori with $\pi_1(T^2)\to \pi_1(S^1\times S^4)$ surjective should be isotopic to the product torus.

For the homotopically trivial case, one might be able to use the technique of the Whitney-Wu unknotting theorem to prove that any two homotopically trivial tori are isotopic; certainly their lifts to the universal cover (or a large enough finite-sheeted cover) will be isotopic.

I think this is true (homotopy implies isotopy). Consider a generic smooth torus embedding $f: T^2 \to S^1\times S^4$, then the projection to $S^1$ gives a circle-valued Morse function on $T^2$. Because of the ambient dimension (codimension 3), I think one may isotope $f$ to cancel critical points and obtain an isotopic embedding $f': T^2\to S^1\times S^4$ such that the projection to $S^1$ has no critical points when restricted to $T^2$ if the map $f_{|T^2}:T^2\to S^1$ is homotopically non-trivial. Thus $f'^{-1}(z)$ is a union of $k$ circles in $S^4$ for all $z\in S^1$, where $k$ is the index of $\pi_1(T^2)\to \pi_1(S^1)$. The homotopy type of $f'$ should be determined by $k$

For the homotopically trivial case, I think the technique of the Whitney-Wu unknotting theorem implies that any two homotopically trivial tori embedded in $S^1\times S^4$ are isotopic. The point is that we may use Morse theory to shrink the torus down into an $I\times S^4$ slice, then apply Whitney-Wu.

Now consider the homotopically non-trivial case, where $f’^{-1}(z)$ consists of $k$ circles for every $z$. We may think of this as a map $g: S^1 \to Emb(\sqcup^k S^1, S^4)$, where the monodromy permutes the components of $\sqcup^k S^1$ cyclically so that the mapping torus is a torus. Then $g(1)$ is an embedding of $\sqcup^k S^1$ to $S^4$. Let $g(1)=\gamma_1 \sqcup \cdots \sqcup \gamma_k$ a disjoint union of circles. By general position, this is the boundary of an embedding $D^2_1 \sqcup \cdots \sqcup D^2_k$. Assume that the monodromy sends $\gamma_i$ to $\gamma_{i+1}$, indices taken $(\mod k)$. We may extend $g$ to an ambient isotopy of $S^4$, and hence get a path $G: [0,1] \to Emb( \sqcup^k D^2 , S^4)$ with boundary restricting to $g$. We may also assume that $D_i$ is taken to $D_{i+1}$ by the isotopy, $1\leq i< k$, and $D_k$ is taken to $D_1’$ with $\partial D_1’=\gamma_1$. 

By the 4-dimensional light-bulb theorem (Theorem 1.10), $D_1$ is isotopic to $D_1'$ rel $\gamma_1$. By general position, we may assume that this isotopy misses $D_2, \ldots, D_k$ (since these all sit inside of disjoint balls).

So this isotopy may be achieved by a 1-parameter isotopy of disks $G’:S^1 \to Emb(\sqcup^k D^2,S^4)$. But any such isotopy is homotopic to a cyclic rotation of order $k$. Consider a point on the disk $D_1$, the point gives a loop in $S^4$ by iterating the isotopy $k$ times, which is contractible, and hence one may assume that the isotopy the points on the disks $D_1, \dots, D_k$ arranged around a round circle. Then one may also assume that the tangent space to the disks at the point are permuted cyclically, and then that the disks are permuted by a rotation by shrinking them down to the tangent space. Hence there should be only one isotopy class of embedded tori for each $k$.

It’s possible that the application of the 4D lightbulb trick here is overkill, one might look at the classical isotopy theory linked to above to see if it can be applied directly to show that homotopy implies isotopy in this case (things tend to be more flexible in higher dimensions).

I think there is a possible reduction of this question. Consider a generic smooth torus embedding $f: T^2 \to S^1\times S^4$, then the projection to $S^1$ gives a Morse function on $T^2$. Because of the ambient dimension (codimension 3), I think one may isotope $f$ to cancel critical points and obtain an isotopic embedding $f': T^2\to S^1\times S^4$ such that the projection to $S^1$ has no critical points when restricted to $T^2$ if the map $f_{|T^2}:T^2\to S^1$ is homotopically non-trivial. Thus $f'^{-1}(z)$ is a union of circles in $S^4$ for all $z\in S^1$. The homotopy type of $f'$ should be determined by the number of circles, or correspondingly the index of the image $\pi_1(T^2)\to \pi_1(S^1)$.

If the configuration space of $k$ circles in $S^4$ has $\pi_1$ isomorphic to the symmetric group $S_k$, then I think that one should be able to show that any two such maps are isotopic.

For $k=1$, ie the case that $f'^{-1}(z)$ has a single circle, I think one can show that all such tori are isotopic. Because this is codimension $3$ in $S^4$ it is unknotted.

Let $f: S^1 \times S^1 \to S^4$ be a map so that $f(z,S^1)$ is embedded for all $z$, then we may think of $f$ as a loop in in the configuration space of embeddings of a circle in $S^4$. This loop may be achieved by a 1-parameter family of diffeomorphisms by ambient isotopy. Let $D$ be an embedded disk bounding $\gamma= f(1,S^1)$, and let $D'$ be the disk obtained after the ambient isotopy. By the 4-dimensional light-bulb theorem (Theorem 1.10), $D$ is isotopic to $D'$ rel $\gamma$. So this isotopy may be achieved by a 1-parameter isotopy of disks. But any such isotopy is homotopic to the constant isotopy. Consider a point on the disk, the point gives a loop in $S^4$ which is contractible, and hence one may assume that the isotopy fixes this point. Then one may also assume that the tangent space to the disk at the point is fixed, and then that the disks are fixed by shrinking them down to the tangent space. Hence I think that the tori with $\pi_1(T^2)\to \pi_1(S^1\times S^4)$ surjective should be isotopic to the product torus.

I think there is a possible reduction of this question. Consider a generic smooth torus embedding $f: T^2 \to S^1\times S^4$, then the projection to $S^1$ gives a Morse function on $T^2$. Because of the ambient dimension (codimension 3), I think one may isotope $f$ to cancel critical points and obtain an isotopic embedding $f': T^2\to S^1\times S^4$ such that the projection to $S^1$ has no critical points when restricted to $T^2$ if the map $f_{|T^2}:T^2\to S^1$ is homotopically non-trivial. Thus $f'^{-1}(z)$ is a union of circles in $S^4$ for all $z\in S^1$. The homotopy type of $f'$ should be determined by the number of circles, or correspondingly the index of the image $\pi_1(T^2)\to \pi_1(S^1)$.

If the configuration space of $k$ circles in $S^4$ has $\pi_1$ isomorphic to the symmetric group $S_k$, then I think that one should be able to show that any two such maps are isotopic.

For $k=1$, ie the case that $f'^{-1}(z)$ has a single circle, I think one can show that all such tori are isotopic. Because this is codimension $3$ in $S^4$ it is unknotted.

Let $f: S^1 \times S^1 \to S^4$ be a map so that $f(z,S^1)$ is embedded for all $z$, then we may think of $f$ as a loop in in the configuration space of embeddings of a circle in $S^4$. This loop may be achieved by a 1-parameter family of diffeomorphisms by ambient isotopy. Let $D$ be an embedded disk bounding $\gamma= f(1,S^1)$, and let $D'$ be the disk obtained after the ambient isotopy. By the 4-dimensional light-bulb theorem (Theorem 1.10), $D$ is isotopic to $D'$ rel $\gamma$. So this isotopy may be achieved by a 1-parameter isotopy of disks. But any such isotopy is homotopic to the constant isotopy. Consider a point on the disk, the point gives a loop in $S^4$ which is contractible, and hence one may assume that the isotopy fixes this point. Then one may also assume that the tangent space to the disk at the point is fixed, and then that the disks are fixed by shrinking them down to the tangent space. Hence I think that the tori with $\pi_1(T^2)\to \pi_1(S^1\times S^4)$ surjective should be isotopic to the product torus.

For the homotopically trivial case, one might be able to use the technique of the Whitney-Wu unknotting theorem to prove that any two homotopically trivial tori are isotopic; certainly their lifts to the universal cover (or a large enough finite-sheeted cover) will be isotopic.