If I'm understanding the question, the answer is pretty much "no", to the best of my knowledge. There is a notion of closed category (note the absence of the word "monoidal") which involves a mixed variance bifunctor, say $[-\Rightarrow -]$, satisfying certain naturality conditions that make it resemble an exponential minus the universal property. The best you can say is that if all functors of the form $[a\Rightarrow-]$ have left adjoints, then the category is also monoidal closed; but what makes for "real" exponentials is exactly that the monoidal structure is the Cartesian monoidal structure. Exponentiation doesn't distinguish itself as a closed structure in any other way that I'm aware of.
The reason that you can seemingly treat function types in theories like HoTT without even having product types is that the syntax of the language smuggles in products automatically. They do this with their context, the list of variables we're considering free in our type expression. The structural rules (as seen in the HoTT book's appendix) tell you in a sly way that the lists of free variables behave like products, and terms-in-context are actually morphisms with domains products. That is, you should think of a judgment like "$x:A,y:B\vdash t:C$" as notation for $t:A\times B\to C$ in whatever category our types might be interpreted in. Lambda abstraction, which consists of passing from the previous sequent to "$x:A\vdash\lambda y.t:B\to C$", can then be seen to be just to be the syntax of the exponential transposition operation (when considered in the company of all the introduction, elimination, and computation rules).