Where is there a treatment of "exponential monads"? I have a category $C$, which is equipped with a symmetric monoidal structure (tensor product $\otimes$, unit object $1$).  My category also has finite coproducts (I'll write them using $\oplus$, and write $0$ for the initial object), and $\otimes$ distributes over $\oplus$.
By an exponential monad, I mean a monad $(T,\eta,\mu)$ on $C$, where the functor $T:C\to C$ is equipped with some structure maps of the form
$$\nu \colon 1 \to T(0)$$
and
$$\alpha\colon T(X)\otimes T(Y) \to T(X\oplus Y).$$
The structure maps are isomorphisms, and are suitably "coherent" with respect to the two monoidal structures $\otimes$ and $\oplus$.
The simplest example is: $C$ is the category of $k$-vector spaces, and $T=\mathrm{Sym}$ is the commutative $k$-algebra monad (i.e., $\mathrm{Sym}(X)$ is the symmetric algebra $\bigoplus \mathrm{Sym}^q(X)$).
Now, I'm sure I can work out all the formalism that I need for this, if I have to.  My question is: is there a convenient place in the literature I can refer to for this?  Alternately, is there suitable categorical language which makes this concept easy to talk about?
I'd also like to have a good formalism for talking about a "grading" on $T$.  This means a decomposition of the functor $T=\bigoplus T^q$, where $T^q\colon C\to C$ are functors, which have "nice" properties (for instance, $T^m(X\oplus Y)$ is a sum of $T^p(X)\otimes T^{m-p}(Y)$).  The motivating example again comes from the symmetric algebra: $\mathrm{Sym}=\bigoplus \mathrm{Sym}^q$.
 A: I don't know if there's a standard name for what you're calling "exponential monads".  Maybe not.  But such things have been considered.  You mention the example of the free commutative algebra monad $\mathrm{Sym}$ on $\mathbf{Vect}$.   The linear structure is inessential here, so you could simplify and consider the free commutative monoid monad $T$ on $\mathbf{Set}$ instead.  Perhaps this deserves to be called the exponential monad, since 
$$
T(X) =
\sum_{n = 0}^\infty X^n/S_n
$$
where $S_n$ is the $n$th symmetric group, which has $n!$ elements.  (Here $\sum$ denotes coproduct.)  I believe there is even a sense in which the derivative of $T$ is $T$.  Your monad $\mathrm{Sym}$ might be called the "linearized exponential monad".   
Exponential monads (and in particular the free commutative monoid monad) have been considered in linear logic.  I'm probably not giving the canonical reference here, but you could try at a paper of Marcelo Fiore, Differential structure in models of multiplicative biadditive intuitionistic linear logic.  For example, on page 8 he mentions 

the Seely monoidal natural isomorphism
  $$s: !A \otimes !B\ \stackrel{\cong}{\to}\ !(A \times B)$$

Here $!$ is a comonad (you might have to do some dualizing), and $\times$ is a "biproduct", i.e. simultaneously a product and a coproduct.  (In your example of $k$-vector spaces, coproducts are biproducts.)  He cites work of Blute, Cockett and Seely on differential categories, which might be worth chasing up in the hope of finding a treatment of exponential monads.  
I think this use of the word "differential" is connected to my dim recollection above, that the derivative of the free commutative monoid monad is itself --- whatever that means. 
Edit You might get a better response if you ask the categories mailing list, categories@mta.ca.  There are certainly people there who know more about this than me.
A: I've just spent some time working out for myself the details of what an exponential monad is supposed to be, so I might as well post what I learned here.  (I don't know that I'll ever have a reason to write it up more formally.)
Here is the correct definition.  Given symmetric monoidal $(C,\otimes, 1)$, with finite coproducts $+$ and initial object $0$, an exponential structure on a monad $T$ on $C$ should be 


*

*the structure of strong symmetric monoidal functor on $T: (C,+,0)\to (C,\otimes, 1)$, consisting of natural isomorphisms $\nu\colon 1\to T0$ and $\alpha\colon TX\otimes TY\to T(X+Y)$ satisfying a bunch of coherence properties.


There is one additional condition you need to impose.  To state this condition, let $\gamma: T(TX\otimes TY)\to TX\otimes TY$ be the composite
$$ T(TX\otimes TY) \xrightarrow{T\alpha} TT(X+Y) \xrightarrow{\mu} T(X+Y) \xrightarrow{\alpha^{-1}} TX\otimes TY,
$$
where $\mu: TT\to T$ is part of the monad structure.  The additional condition is 


*

*for all $X$ and $Y$, we have $\gamma\circ T(\mu\otimes \mu)=(\mu\otimes \mu)\circ \gamma$ as maps $T(TTX\otimes TTY)\to TX\otimes TY$.


The map $\gamma$ defines a $T$-algebra structure on $TX\otimes TY$ (it is the free algebra structure on $T(X+Y)$, transported along the isomorphism $\alpha$); the additional property says that $\mu\otimes \mu$ is itself a map of $T$-algebras.  Note also that the map $\nu$ identifies $1$ with the initial $T$-algebra $T0$.
Given this, you can prove (with no more difficulty than you would expect) that for every pair of $T$-algebras $A$ and $B$, we can put a canonical $T$-algebra structure on $A\otimes B$, which exhibits it as the coproduct of $A$ and $B$ in the category of $T$-algebras.  In particular, the forgetful functor from $T$-algebras to $C$ becomes strong symmetric monoidal, using coproduct as the monoidal structure for $T$-algebras.  
I would have expected that you would need another condition, relating $\gamma$ to the unit map $\eta\colon I\to T$, or perhaps a condition on $\nu$, but it doesn't seem that this is necessary as far as I can tell.
You don't need the hypothesis that $\otimes$ distribute over coproduct, as I suggested in my question.  It might seem surprising, but apparently you don't even need the monoidal structure on $C$ to be symmetric or associative; being unital appears to be enough to exhibit coproducts of $T$-algebras using $\otimes$.  I suspect that for such a "unital monoidal" category $C$, you might be able to show that the existence of an exponential monad implies that $C$ is symmetric monoidal.  (This does not seem so crazy in light of the way the symmetric monoidal smash product of  EKMM spectra comes about).
