The answer is no in general. Actually, the opposite inequality is true!
The inequality fails when all the primes of minimal height containing $I$ are destroyed by the localization. Here is an example:
Let $k$ be a field and let $R=k[x_1,x_2,x_3,\dots]$. (The countable sequence of indeterminates is to satisfy the OP's requirement that $R$ be non-noetherian, but actually the example works with $R=k[x_1,x_2,x_3]$.) Let $S$ be generated by $x_1$ (i.e. $S=\{1,x_1,x_1^2,\dots\}$). Let $I = (x_1x_2,x_1x_3)$.
Then $\operatorname{ht}(I) = 1$, because $I\subset (x_1)$, which is of height $1$. However, $\operatorname{ht}(I^e)=2$, because $I^e = (x_2,x_3)\supset (x_2)\supset 0$.
I assume the question is based on the intuition that because $S^{-1}R$ has fewer primes than $R$, heights ought to be lower, so perhaps it is counterintuitive that actually, in all cases, $\operatorname{ht}(I)\leq \operatorname{ht}(I^e)$, but I will argue that this is actually what's true of all commutative, unital rings $R$ and all multiplicatively closed subsets $S$ (at least with the convention that the height of the unit ideal is $+\infty$).
Lemma: Let $R$ be a commutative, unital ring, $S$ a multiplicatively closed subset, $I$ an ideal of $R$, $\mathfrak{p}$ a prime ideal of $R$ that does not meet $S$, and suppose that $I^e \subset \mathfrak{p}^e$. Then $I\subset \mathfrak{p}$.
Proof: $I^e\subset \mathfrak{p}^e$ means that for any $a/s\in I^e$ (with $a\in A$ and $s\in S$), there is a $p/s'\in \mathfrak{p}^e$ with $a/s = p/s'$ in $S^{-1}R$, i.e. there is an $s_0\in S$ with $s_0(s'a - sp) = 0$ in $R$. Since $S$ does not meet $\mathfrak{p}$, we know that $s_0\notin \mathfrak{p}$, so we have $s'a - sp \in\mathfrak{p}$ because $\mathfrak{p}$ is prime, and therefore $s'a\in\mathfrak{p}$. Again, $s'\notin \mathfrak{p}$, and since $\mathfrak{p}$ is prime, we conclude $a\in\mathfrak{p}$. Applying this to $a/1\in I^e$ for any $a\in I$, we conclude that $I\subset\mathfrak{p}$.
Proposition: For any commutative, unital ring $R$, any ideal $I$ of $R$ of finite height, and any multiplicative set $S\subset R$, we have
$$\operatorname{ht}(I)\leq \operatorname{ht}(I^e),$$
where $I^e$ is the extension of $I$ in $S^{-1}R$.
Proof:
Case 1: $S$ meets $I$. Then $I^e$ is the unit ideal, and its height is $+\infty$, so the desired inequality is trivial in this case.
Case 2: $S$ does not meet $I$. If $I^e$ does not have finite height, the required inequality is true, so we can assume it has finite height $r$. Let $\mathfrak{P}_0\subset\mathfrak{P}_1\subset\dots\subset\mathfrak{P}_r$ be a saturated chain of primes of $S^{-1}R$ of minimal length such that $I^e\subset \mathfrak{P}_r$.
The prime ideals of $S^{-1}R$ are in bijection with the prime ideals of $R$ not meeting $S$ (e.g. Atiyah-MacDonald proposition 3.11), with the bijection given by $\mathfrak{p}\triangleleft R\mapsto \mathfrak{p}^e\triangleleft S^{-1}R$. So, for each $\mathfrak{P}_j$, there is a prime $\mathfrak{p}_j\triangleleft R$ such that $\mathfrak{p}_j^e= \mathfrak{P}_j$, and none of the $\mathfrak{p}_j$'s meet $S$. Then the lemma implies $\mathfrak{p}_0\subset\mathfrak{p_1}\subset\dots\subset\mathfrak{p}_r$, and also $I\subset\mathfrak{p}_r$. Since the chain $\mathfrak{P}_0\subset\dots\subset\mathfrak{P}_r$ is saturated, the chain $\mathfrak{p}_0\subset\dots\subset\mathfrak{p}_r$ must also be saturated, because any $\mathfrak{p}'$ with $\mathfrak{p}_j\subset\mathfrak{p}'\subset\mathfrak{p}_{j+1}$ would avoid $S$ (since $\mathfrak{p}_{j+1}$ does), and $\mathfrak{p}\mapsto \mathfrak{p}^e$ is injective and order preserving on primes that avoid $S$, so it would imply a $\mathfrak{P}'$ between $\mathfrak{P}_j$ and $\mathfrak{P}_{j+1}$, contrary to the saturation assumption. So we have exhibited a saturated chain of length $r$ with the top prime $\mathfrak{p}_r$ containing $I$, and we can conclude $\operatorname{ht}(I)\leq r = \operatorname{ht}(I^e)$ as desired.
