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Source. Section 6, printed p. 152 (published PDF).

Statement. Let $S$ be a nonempty subset of a finite matroid. The following conditions are equivalent:

1. Either every element of $S$ is a coloop, or no element of $S$ is a coloop and every base omits at most one element of $S$.
2. All elements of $S$ belong to exactly the same circuits.

This is the source's meaning of elements being in series. The source uses isolated for what is called a coloop here.

Proof. Assume condition 2. If no circuit contains any element of $S$, all its elements are coloops by the circuit criterion. Otherwise a circuit containing one contains them all, so none is a coloop. If a base $B$ omitted two distinct elements $x,y\in S$, then $B\cup\{x\}$ would contain a circuit containing $x$ but not $y$, contrary to condition 2. Thus condition 1 follows.

Conversely, if all elements of $S$ are coloops, they all belong to no circuits, giving condition 2. Consider the other case of condition 1. If condition 2 fails, there are $x,y\in S$ and a circuit $C$ with $x\in C$ and $y\notin C$.

Because $y$ is not a coloop, $r(E\setminus\{y\})=r(E)$. The circuit $C$ remains a circuit in the restriction to $E\setminus\{y\}$. Consequently $x$ is not a coloop of that restriction, and the same rank criterion gives

$$r(E\setminus\{x,y\}) =r(E\setminus\{y\})=r(E).$$

A base of the double deletion is therefore a base of the original matroid omitting both $x$ and $y$. This contradicts condition 1, and proves the equivalence. $\square$