The central facts governing the relation between generic filters arising from different forcing notions are the following:
Theorem. Suppose that $G\subset\mathbb{P}$ and $H\subset\mathbb{Q}$ are $V$-generic.
If $V[G]=V[H]$, then there are conditions $p\in G$ and $h\in H$ such that $\text{RO}(\mathbb{P}\upharpoonright p)\cong\text{RO}(\mathbb{Q}\upharpoonright h)$.
If $V[G]\subset V[H]$, then there are conditions $p\in G$ and $h\in H$ such that $\text{RO}(\mathbb{P}\upharpoonright p)$ is isomorphic to a complete subalgebra of $\text{RO}(\mathbb{Q}\upharpoonright h)$.
It follows that if $c$ is a Cohen real, then it isn't $V$-generic for any forcing notion, except those that look just like Cohen forcing below a condition. Other forcing notions, of course, can add Cohen reals as a by-product, and this often happens at limit stages in finite support iterations or products of forcing. In these cases, these other forcing notions have Cohen forcing as a complete subalgebra, and they can be viewed as adding a Cohen real, followed by the corresponding quotient forcing.
I take the first part of the theorem to say basically that a generic filter determines in a sense the forcing notion that gave rise to it, and that a generic filter can't by accident be generic for a totally different forcing notion.
I can post a proof sketch of the theorem, if you like. For the first claim, let $\dot H$ be a $\mathbb{P}$-name, and then let $p\in G$ force that $\dot H$ is $\check V$-generic and that $\check V[\dot G]=\check V[\dot H]$. One then looks at the map $q\mapsto [\![q\in\dot H]\!]^{\mathbb{P}\upharpoonright p}$.
One must restrict to the forcing below a condittion in this theorem, because otherwise one can do silly sneaky stuff like this: let $\mathbb{B}$ be Cohen forcing, and let $\mathbb{C}$ be the lottery sum of trivial forcing with Cohen forcing. So if $c$ is $V$-generic for $\mathbb{B}$, then it can also be added by the second, by opting for the part of the forcing the selects Cohen forcing; but $\mathbb{B}$ does not embed densely into $\mathbb{C}$, because $\mathbb{C}$ has an atom, the part that opted instead for trivial forcing, and there is no way to embed anything below that atom. So this technically counts as an affirmative answer to your highlighted question. But what it shows is that really one wants instead to consider cones in the forcing.
