Substitute your central vertex in your graph with a 3-cycle $abc$ so that the graph stays cubic. Now subdivide each edge in this 3-cycle. So we have new vertices $u$ connected to $a$ and $b$, $v$ connected to $b$ and $c$, $w$ connected to $c$ and $a$. Now add a final vertex $x$ and connect it to $u,v$ and $w$. This graph has exactly three bridges, none of which intersect the other at a vertex, and moreover has no perfect matching!

One result which relates the existence of a perfect matching in a cubic graph and its bridges is the following theorem of Petersen from "Die theorie der regularen graphen", Acta Math. 15
(1891), 163-220:

**Theorem**: Every cubic graph with at most two bridges contains a perfect matching.

As well as this strengthening by Errera, "Du colorage des cartes", Mathesis 36 (1922), 56-60:

**Theorem**: If all the bridges of a connected cubic graph $G$ lie on a single path of $G$, then $G$ has a perfect matching.

So your instinct is true, in the sense that if the graph has no perfect matching, its bridges do not lie on a path. However the example in the beginning of this answer shows that they are not necessarily incident at the same vertex.

smallest number of vertices among all cubic, edge-1-connected graphs without a perfect matching. A proof for this statement was published in Gary Chartrand, Donald L. Goldsmith, Seymour Schuster:A sufficient condition for graphs with 1-factors. Colloquium Mathematicum. Volume XLI, Fascicle 2, 1979. See Corollary 2a on p. 343 for the statement, and p. 341 for a picture. $\endgroup$ – Peter Heinig Sep 20 '17 at 15:32