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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 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, for any $g\in{\mathcal S}({\mathbb R})$ there is an $\varepsilon>0$ such that $\varepsilon\cdot B\subseteq\{f\in{\mathcal S}({\mathbb R}):\ |\langle f,g\rangle|\le 1\}$, 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 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, ${\mathcal S}'({\mathbb R})$ 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, 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 totally bounded sets in ${\mathcal S}({\mathbb R})$), where the same reasoning works also.

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, ${\mathcal S}'({\mathbb R})$ 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, 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.

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 sequences; on the other hand, by the Banach-Steinhaus theorem, ${\mathcal S}'({\mathbb R})$ 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, 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 totally bounded sets in ${\mathcal S}({\mathbb R})$), where the same reasoning works also.

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 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, for any $g\in{\mathcal S}({\mathbb R})$ there is an $\varepsilon>0$ such that $\varepsilon\cdot B\subseteq\{f\in{\mathcal S}({\mathbb R}):\ |\langle f,g\rangle|\le 1\}$, 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 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, ${\mathcal S}'({\mathbb R})$ 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, 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 totally bounded sets in ${\mathcal S}({\mathbb R})$), where the same reasoning works also.

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 sequences; on the other hand, by the Banach-Steinhaus theorem, ${\mathcal S}'({\mathbb R})$ 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, 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 reasonong works also.

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 sequences; on the other hand, by the Banach-Steinhaus theorem, ${\mathcal S}'({\mathbb R})$ 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, 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 totally bounded sets in ${\mathcal S}({\mathbb R})$), where the same reasoning works also.