Suppose we have two closed oriented 3-manifolds $M$ and $N$. Suppose $N$ is obtained by a Dehn surgery operation on a knot $K$ in $M$, so $N=(M-\operatorname{int}\nu K)\cup_\partial (S^1\times D^2)$, with coefficient $p/q$, so that the meridian of $S^1\times D^2$ corresponds to $p \mu + q\lambda$ where $\mu$ is a right-handed meridian of $K$ and $\lambda$ is a parallel copy of $K$. (Here $\nu K$ is a closed tubular neighborhood of $K$.)
Is it true that we can say that $M$ can be obtained by a Dehn surgery operation from $N$?
Is my following argument correct? Let $K'=S^1\times \{0\} \subset S^1\times D^2 \subset (M-\operatorname{int}\nu K)\cup_\partial (S^1\times D^2) =N$. Then $\nu K'=S^1\times D^2$ and $N-\operatorname{int} \nu K'=M-\operatorname{int}\nu K$, so $M=(N-\operatorname{int} \nu K')\cup _\partial (S^1\times D^2)$ for some suitable boundary homeomorphism. This means that $M$ is obtained by a Dehn surgery on $K'\subset N$.
Also can we determine the surgery coefficient of $K'$?
Yes, you can say that $M$ is obtained by a Dehn surgery operation from $N$, and I think that your argument is basically correct.
But when asking whether one can determine "the" surgery coefficient of $K'$, there is a conceptual issue to deal with, in that $\lambda$ is not well-defined, and therefore surgery coefficients are not well defined.
What does $\lambda$ is a parallel copy of $K$ even mean?
If it means $\lambda$ is a simple closed curve in $\partial(\nu K)$ which is isotopic to $K$ in $\nu K$, the problem is that there are infinitely many choices of $\lambda$ up to isotopy in the torus $\partial(\nu K)$. These choices differ from each other by powers of the Dehn twist around $\mu$, and these choices affect the surgery coefficient.
In knot theory, where $M=S^3$, by imposing an additional constraint it is possible to pin down a particular choice of $\lambda$ uniquely up to isotopy in $\partial(\nu K)$: one requires that $\lambda$ represents the trivial element of the homology group $H_1(M - \text{int}(\nu K)) \approx \mathbb Z$. There is a similarly well-defined choice of $\lambda$ in any situation where $M$ is a homology 3-sphere (and so, in such situations, surgery coefficients are well-defined). But in general there is no similar constraint which lets one pin down a well-defined choice of $\lambda$.
Perhaps on the other hand you are willing to choose an arbitrary homeomorphism $S^1 \times S^1 \mapsto \partial(\nu K)$, subject perhaps to the "meridian constraint" saying that $\text{(point)} \times S^1$ corresponds to $\mu$. Having made that choice, now you can choose $\lambda$ to correspond to $S^1 \times \text{(point)}$. And now you get surgery coefficients. And then sure, you can determine the surgery coefficient for $K'$ just as is done in knot theory.
But one should keep in mind two things: all of this depends on the choice of the original homeomorphism $S^1 \times S^1 \mapsto \partial(\nu K)$; and (up to isotopy in $\partial(\nu K)$) that choice is pretty much equivalent to the choice of $\lambda$ itself.