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

To tell the truth, 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 the topology on ${\mathcal S}({\mathbb R})$ which is stronger than the weak topology generated on ${\mathcal S}({\mathbb R})$ by the duality
$$
\langle f,g\rangle=\int_{\mathbb R}f(t)\cdot g(t)\, d t
$$
(in other words, for any $g\subseteq{\mathcal S}({\mathbb R})$ there is an $\varepsilon>0$ such that $\varepsilon\cdot B\subseteq\{f:\ |\langle f,g\rangle|\le 1\}$, where $B$ is again the unit ball of $\|\cdot\|$).

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 after taking the completion it turns 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 seqiences; 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 usual topology on ${\mathcal S}'({\mathbb R})$ wich is usually called strong (i.e. the topology of uniform convergence on bounded or, what is the same here, on compact sets in ${\mathcal S}({\mathbb R})$), where the same construction 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