I am not sure that this is what you want, but it's too long for a comment, so I post it as an answer. 

I don't see serious problems. If the conditions in terms of the theory of topological vector spaces will satisfy you, then you can consider the following. Your norm $\|\cdot\|$ must 

 1. be continuous on ${\mathcal S}({\mathbb R})$ (in other words, the
    unit ball $B$ of this norm must be a neighbourhood of zero in
    ${\mathcal S}({\mathbb R})$), and
    
 2. generate a topology on ${\mathcal S}({\mathbb R})$ which is
    stronger than the weak topology generated on ${\mathcal S}({\mathbb
    R})$ by the duality $$ \langle f,g\rangle=\int_{\mathbb R}f(t)\cdot
    g(t)\, d t,\quad f,g\in{\mathcal S}({\mathbb R}) $$ (in other words,
    $\forall g\in{\mathcal S}({\mathbb R})\ \sup_{f\in{\mathcal
    S}({\mathbb R}):\ \|f\|\le 1}|\langle f,g\rangle|<\infty$).

If you denote by ${\mathcal S}({\mathbb R})_{\|\cdot\|}$ the space ${\mathcal S}({\mathbb R})$ endowed with the topology generated by such a norm, then you obtain a chain of continuous mappings
$$
{\mathcal S}({\mathbb R})\subseteq{\mathcal S}({\mathbb R})_{\|\cdot\|}\subseteq {\mathcal S}'({\mathbb R}),
$$
and the completion turns it into the chain 
$$
{\mathcal S}({\mathbb R})\subseteq X\to {\mathcal S}'({\mathbb R}).
$$
(since ${\mathcal S}({\mathbb R})_{\|\cdot\|}$ is normed, its completion consists of Cauchy sequences; on the other hand, by the [Banach-Steinhaus theorem][1], ${\mathcal S}'({\mathbb R})$ is sequentially complete with respect to the ${\mathcal S}({\mathbb R})$-weak topology, so the images of these sequences have limits in ${\mathcal S}'({\mathbb R})$).

The only detail which is not clear here is if the last mapping (the arrow $\to$) is injective. In general case [completion does not preserve injectivity][2], so you should verify this in case that this is important for you. 

You can also consider the strong topology on ${\mathcal S}'({\mathbb R})$ (i.e. the topology of uniform convergence on bounded or, what is the same here, on totally bounded sets in ${\mathcal S}({\mathbb R})$), where the same reasoning works also.


  [1]: https://en.wikipedia.org/wiki/Uniform_boundedness_principle
  [2]: https://math.stackexchange.com/questions/1744400/does-the-operation-of-completion-preserve-injectivity