No, there is not tensor product. You need to make further assumptions that could give you several potentially distinct tensor products.

You can always form the product category $M\times N$. It is a monoidal category. Any sensible tensor product will be a quotient category of $M\times N$.

For instance, in the Deligne tensor product, mentioned by McRae, you assume that $M$ and $N$ are abelian, $k$-linear. Then $M\times N$ is $k$-bilinear and you can ask for the universal quotient $M\otimes_k N$ such that any $k$-bilinear monoidal functor $M\times N\rightarrow P$ factors via a $k$-linear monoidal functor $M\otimes_k N\rightarrow P$. Such thing clearly exists: you dont even need to assume that the categories are abelian but you need $k$-linearity! However, $M\otimes_k N$ is not necessarily abelian, even if $M$ and $N$ are such. If it is abelian, then it is Deligne tensor product.

Overall, you need some "balancing" conditions to form a tensor product. In the case of $k$-linear categories, the balancing is performed by $Vec(k)$, the category of vector spaces that acts on both $M$ and $N$. There are several ways you can do "balancing" but only some of them preserve monoidal and/or braided structure. Others will lose it.

For instance, the following wabbity balancing is cool. For this I need a third monoidal category $C$ and two monoidal functors $C\rightarrow M$ and $C\rightarrow N$. By a $C$-balanced functor I understand a bifunctor $M\times N \rightarrow D$, together with an equivalence of two resulting trifunctors $M\times C \times N \rightarrow D$. I claim that there exists a monoidal category $M\otimes_CN$ with the universal $C$-balanced bifunctor $M\times N \rightarrow M\otimes_C N$, with two caveats:

- It is not longer braided, you need some conditions for that.
- It is no longer a category :-)) The issue is that your hom-s will be proper classes, unless you put some smallness condition, for instance, $C$ being small will do.