Does there exist an algebraic group $G$ such that all semisimple elements with infinite order lie in $G^0$? Let $G$ be a linear algebraic group over an algebraically closed field $k$.  If $G$ is connected, and $1_G \neq s \in G$ is semisimple, then $s$ lies in a nontrivial torus: this is Proposition 6.4.5 in Springer, Linear Algebraic Groups.
Now assume $G$ is not connected, and let $1_G \neq s \in G$ be semisimple.
The identity component $G^0$ is normal and of finite index $n = [G :
G^0]$ in $G$, so $s^n \in G^0$ is semisimple, and as long as $1_G \neq
s^n$, we can conclude that $s^n$ lies in a nontrivial torus.
But is it possible to have $s^n = 1_G$ for all semisimple $s \in G - G^0$?
 A: The direct product $\mathbb Z/p\mathbb Z \times \mathrm{GL}_1$ (no need for the second factor, but you wanted a positive-dimensional example), where I abuse notation by using the same notation $\mathbb Z/p\mathbb Z$ for a finite abstract group and the associated constant group variety, is an example of the desired sort when working over a field of positive characteristic $p$, because all semisimple elements lie in the identity component.
In any characteristic, the direct product $\mathbb Z/n\mathbb Z \times \mathfrak{gl}_1$ (where $\mathfrak{gl}_1 = \operatorname{Spec}(k[x])$ is what is sometimes called `Add') has the desired property, for essentially the opposite reason.  (This is basically @MattSamuel's example.)
A: My previous answer may fairly be regarded as somewhat trivial:  it satisfies your stated conditions, but, probably, it doesn't 'feel' like what you want.  There's a good reason for that.  TL;DR:  it can only happen in a 'non-trivial' way if $G^\circ$ is soluble.
Namely, let $T$ be a maximal torus in $\mathrm C_G(s)$.  All geometric points of the coset $s T$ are semisimple, and, unless $T$ is trivial, not all geometric points of the coset $s T$ can have finite order.  (If $s$ has infinite order, then there you go; otherwise, we may simply choose an infinite-order geometric point in $T$.)  For an algebraic group to contain no non-trivial maximal torus is quite a strong condition; it is equivalent to saying that its identity component is unipotent.
Write $R$ for the unipotent radical of $G$.  Corollary 9.4 of Steinberg's wonderful paper "Endomorphisms of linear algebraic groups" shows that $\mathrm C_{G/R}(s)^\circ$ is reductive.  Suppose for the moment that it is not the trivial group.  Then it possesses a non-trivial maximal torus $T'$, which is the image in $G/R$ of some (necessarily non-trivial) maximal torus in $G$.  (This is true in general for morphisms of algebraic groups, by Proposition 11.14(1) of Borel's "Linear algebraic groups".)
Thus, the only way the situation that you describe can arise is if, for all semisimple elements of $G$ not belonging to its identity component, the fixed-point group $\mathrm C_{G/R}(s)$ is finite.  By Corollary 10.12 of Steinberg's paper, this implies that either there are no semisimple elements of $G$ outside the identity component; or $G^\circ/R$, hence $G^\circ$ itself, is soluble.
Certainly there are examples in this last situation; for example, if $G = \mathbb Z/2\mathbb Z \ltimes \operatorname{GL}_1$, with the non-trivial element of $\mathbb Z/2\mathbb Z$ acting by inversion, then every element outside the identity component is an involution.
