For forcing notions $\mathbb{P}$ and $\mathbb{Q}$ let us write $\mathbb{P}\triangleleft\mathbb{Q}$ if forcing with $\mathbb{Q}$ always adds a $\mathbb{P}$-generic filter over $V$. In other words, $\mathbb{P}\triangleleft\mathbb{Q}$ holds if there is a $\mathbb{Q}$-name $\tau$ such that $\tau[H]$ is a $\mathbb{P}$-generic filter over $V$ whenever $H$ is a $\mathbb{Q}$-generic filter over $V$. If $\mathbb{P}\triangleleft\mathbb{Q}$ and $\mathbb{Q}\triangleleft\mathbb{P}$ need $\mathbb{P}$ and $\mathbb{Q}$ be forcing isomorphic, ie must they produce the same generic extensions? I expect not, but haven't been able to think of a counterexample.
I believe this question can be phrased in terms of Boolean algebras; ie if two complete Boolean algebras are each isomorphic to a complete Boolean subalgebra of the other, need they be forcing isomorphic?
(Incidentally, if the generics the forcings add are done in a 'reversible' way, ie if $\tau,\sigma$ are the $\mathbb{Q},\mathbb{P}$ names for the added generics and always $\tau[\sigma[G]]=G$ and $\sigma[\tau[H]]=H$ then the forcing notions will be equivalent. This is mentioned in the beginning of Shelah's text Proper Forcing).
Best Answer
Here is a counterexample, which works in ZFC without any additional large cardinal or other extra hypothesis. This argument, which verifies the guess I made in my original answer, is the result of a conversation I had with Arthur Apter.
The example involves the forcing $\mathbb{S}$ to add a stationary non-reflecting subset of $\omega_2$, that is, a stationary set $S\subset\omega_2$, such that $S\cap\gamma$ is not stationary for any $\gamma\lt\omega_2$ of cofinality $\omega_1$. Conditions in $\mathbb{S}$ consist of bounded sets $s\subset\omega_2$ satisfying the condition for all $\gamma\leq\sup(s)$. The forcing is $\lt\omega_2$-strategically closed, since in the game where players play a game of length $\omega_2$, with player II playing at limit stages, player II may invent an imaginary club set $c$ which is extended as play proceeds, and she ensures that this club remains disjoint from the conditions $s$ that are played. Thus, the forcing adds no new subsets of $\omega_1$ and in particular, preserves $\omega_2$. Also, it follows that the generic set $S\subset\omega_2$ added by $\mathbb{S}$ is indeed stationary. Note that $\mathbb{S}$ has no $\leq\omega_1$-closed dense subset, since with such a highly closed dense subset we would be able to construct an initial segment of $S$ that contains a club of order type $\omega_1$, which would violate the non-reflecting property.
Next, let $\mathbb{T}$ be the forcing to destroy the stationarity of the set $S$ added by $\mathbb{S}$, by adding a club set $C\subset\omega_2$ with $S\cap C=\emptyset$, using closed initial segments. A bootstrap argument shows that the combined forcing $\mathbb{S}\ast\mathbb{T}$ has a dense subset that consists essentially of $(s,c)$, where $s\subset\gamma=\sup(s)$ and $c$ is a closed set containing $\gamma=\sup(c)$ with $s\cap c=\emptyset$. This dense set is $\leq\omega_1$-closed, and thus the combined forcing $\mathbb{S}\ast\mathbb{T}$ is forcing equivalent to $\text{Add}(\omega_2,1)$. So the situation is that $\mathbb{S}$ makes the regrettable faux pas of creating a stationary non-reflecting set, but $\mathbb{T}$ apologizes, and the combination $\mathbb{S}\ast\mathbb{T}$ is completely mild.
So now we can build the counterexample to Cantor-Bernstein for forcing. Let $\mathbb{P}=\text{Add}(\omega_2,1)$, and let $\mathbb{Q}=\mathbb{P}\ast\mathbb{S}$. Clearly $\mathbb{P}$ is a factor of $\mathbb{Q}$, and $\mathbb{Q}$ is a factor of $\mathbb{P}$, precisely because $\mathbb{Q}\ast\mathbb{T}=\text{Add}(\omega_2,1)\ast\mathbb{S}\ast\mathbb{T}=\text{Add}(\omega_2,1)\ast\text{Add}(\omega_2,1)\cong\text{Add}(\omega_2,1)=\mathbb{P}$. So each embeds completely into the other, but they are not forcing equivalent, because $\mathbb{P}$ has a $\leq\omega_1$-closed dense subset, but $\mathbb{Q}$ does not, and this is a property preserved by forcing equivalence.
The argument easily generalizes to higher cardinals than $\omega_2$ (but not for $\omega_1$).
Here is the original answer:
Here is a counterexample, but it uses a large cardinal. I expect that we will be able to eliminate the large cardinal, perhaps by constructing a similar example down low.
Suppose that $\kappa$ is weakly compact. Let $\mathbb{P}=\text{Add}(\kappa,1)$ be the forcing to add a Cohen subset of $\kappa$ by initial segment. Let $\mathbb{Q}=\text{Add}(\kappa,1)*\mathbb{T}$ be the forcing that first adds a Cohen subset to $\kappa$, and then forces to create a $\kappa$-Suslin tree.
Clearly, $\mathbb{P}$ is explicitly a forcing factor of $\mathbb{Q}$. For the converse direction, observe that the forcing $\mathbb{T}$ to create the $\kappa$-Suslin tree can be followed by the forcing that destroys this Suslin tree $T$, by forcing $\mathbb{D}$ to cover it with $\kappa$-many branches. The combined forcing $\mathbb{T}\ast\mathbb{D}$ is actually isomorphic to the forcing consisting of trees of height less than $\kappa$ that are already covered by the branches. This forcing is ${\lt}\kappa$-closed and hence isomorphic to $\text{Add}(\kappa,1)$. It follows that $\mathbb{Q}\ast\mathbb{D}$ is the same as $\mathbb{P}\ast\mathbb{T}\ast\mathbb{D}$, which is the same as $\text{Add}(\kappa,1)\ast\text{Add}(\kappa,1)$, which is forcing equivalent to $\text{Add}(\kappa,1)$, which is $\mathbb{P}$. Thus, we have argued that a further forcing extension of $\mathbb{Q}$ is isomorphic to $\mathbb{P}$, and so $\mathbb{Q}$ is a factor of $\mathbb{P}$.
But the forcing notions $\mathbb{P}$ and $\mathbb{Q}$ are not always equivalent. For example, it is possible to make the weak compactness of $\kappa$ indestructible by $\text{Add}(\kappa,1)$, that is, by $\mathbb{P}$, but the weak compactness of $\kappa$ is always destroyed by $\mathbb{Q}$, since this adds a $\kappa$-Suslin tree, which is incompatible with $\kappa$ being weakly compact.
This forcing was the basis of Kunen's argument that weak compactness is not downwards absolute. In the forcing extension where the $\kappa$-Suslin tree is created, $\kappa$ is not weakly compact, but the weak compactness is recovered once one destroys the tree, since the combined forcing is just $\text{Add}(\kappa,1)$.
By making the preparatory forcing part of $\mathbb{P}$ and $\mathbb{Q}$, one can show that whenever $\kappa$ is weakly compact, then there are forcing notions that are factors of each other, but not forcing equivalent.
I suspect that a similar example can be made down low at the level of $\omega_2$, but I have to think it through. (If someone else can do this, please post an answer.)