Two provide a little bit of insight, assume for simplicity that the team who wins two (not three) games wins the series.
Simplification
Since you have three games with two possible outcomes each, it is enough to consider a probability with $2^3=8$ elements. Namely
$$
\Omega = \{\omega_{AAA}, \ \omega_{AAB}, \ \omega_{ABA}, \ \dots, \ \omega_{BBB}\}
$$
For example, $\omega_{ABB}$ denotes the event that $A$ wins the first game and $B$ wins the second and third. Now define a random variable $X$ like this:
$$
X=\begin{cases}1, & \text{A wins the series}, \\0, & \text{B wins the series} \end{cases}
$$
Using the notation from above, we see that
$$
X(\omega) =\begin{cases} 1,& \omega \in \{\omega_{AAB}, \ \omega_{ABA},\ \omega_{BAA}, \ \omega_{AAA} \}=:\Omega_A \\ 0, & \omega \in \{\omega_{BBA},\ \omega_{BAB},\ \omega_{ABB}, \ \omega_{BBB} \}:=\Omega_B\end{cases}
$$
Finally we can calculate the desired probability, therefore we define the set of all the $\omega$, where $A$ wins the first game:
$$
C:=\{\omega_{AAA}, \ \omega_{AAB},\ \omega_{ABA}, \ \omega_{ABB} \}.
$$
And now:
$$
\mathbb{P}(\text{A wins series} \:| \text{A wins first game})= \mathbb{P}(X=1\: |\: C) = \mathbb{P}(\Omega_A\: |\: C) =\frac{\mathbb{P}(C\cap \Omega_A)}{\mathbb{P}(C)} = \\
\frac{\mathbb{P}(\{\omega_{AAB}, \ \omega_{ABA},\ \omega_{AAA}\ \})}{\mathbb{P}(C)},
$$
which can be calculated easily using the additivity of $\mathbb{P}$. For example
$$
\mathbb{P}(C) =\mathbb{P}(\omega_{AAA})+\mathbb{P}(\omega_{AAB})+\mathbb{P}(\omega_{ABA})+\mathbb{P}(\omega_{ABB})=\\
p^3+p^2(1-p)+p^2(1-p)+p(1-p)^2.
$$
General Case
For the general case you need to consider a probability space with $2^5 =32$ elements. Then $\Omega$ looks like this:
$$
\Omega= \{\omega_{AAAAA}, \ \omega_{AAAAB}, \dots, \ \omega_{BBBBB}\}.
$$
The interpretation of the $\omega$s is the same as before: e.g., $\omega_{AABBA}$ describes the event that $A$ wins the first two games and the last one, and $B$ wins games 3 and 4.
The random variable $X$ and the set $C$ can also be defined as before: the definition of $X$ yields a partition of $\Omega$ into two set $\Omega_A$ and $\Omega_B$, where $\Omega_A$ contains all events where $A$ wins more often than $B$ and $\Omega_B$ contains all events where $B$ wins more often than $A$.
$C$ is again the set of all events where $A$ wins the first game.
$$
C=\{\omega_{AAAAA}, \ \omega_{AAAAB}, \dots, \ \omega_{ABBBB}\}
$$
And therefore
$$
\mathbb{P}(\text{A wins series} \:| \text{A wins first game})= \mathbb{P}(X=1\: |\: C) = \mathbb{P}(\Omega_A\: |\: C) =\frac{\mathbb{P}(C\cap \Omega_A)}{\mathbb{P}(C)}
$$
Final Coment
Finally, I want to say that for exercises like this, it is way faster to compute the desired probabilty directly, without going into to much detail about the probability space in the shadows. But it is a nice exercise after all.
There is a more general version how to calculate it via thinking about 'states'. Let me illustrate this for your example. Let state be $(a,b)$ where $a$ is the number of games won by $A$ and $b$ is the number of games won by $B$. Value of state $(a,b)$, $v_{a,b}$ is the probability of $A$ wining the series. Hence $v_{3,3}=0.55$ (you can think of $v_{4,3}=1$ and $v_{3,4}=0$ and about the transition probabilities in the next sentence). From this, you work backwards since any state $(a,b)$ transitions either to state $(a+1,b)$ with probability $0.55$ or to state $(a,b+1)$ with probability $0.45$. Hence $$\begin{aligned}
v_{3,2}&=0.55\cdot1+0.45\cdot v_{3,3}=0.7975\\
v_{3,1}&=0.55\cdot1+0.45\cdot v_{3,2}=0.908875\\
\end{aligned}$$ so that your answer is correct. By calculating the value of each state you figure out the probability of $A$ winning for any combination of games already won.
Best Answer
One approach might be to assume that you have a Markov Chain, with steady state probabilities for the last two matches having been WW, WL, LW, LL. Each of these states could have a probability of winning the next match, so giving you a transition matrix. You could then find expressions for the information in the question. Then you need to solve for the eight unknowns (the four steady state probabilities and the four probabilities of winning the next match) subject to constraints.
I suspect this may not be possible with the particular data you give: I think you may have eight unknowns and nine constraints (the four numbers in the question, preserving the four steady state probabilities over time, and the steady state probabilities needing to add up to $1$).
As an example of the problem, "if a football team won the last match then the probability that the team will win the next match is 0.6, and if the team lost the last match, then the probability of winning the next game is 0.5" implies for a steady state that the probability the team won the last match was $\frac59$ and that it lost was $\frac49$, so the probability it wins the next match is $\frac59 \times 0.6 + \frac49 \times 0.5 = \frac59$, so ensuring a steady state. But the probability it wins the match after next is $\frac59 \times 0.55 + \frac49 \times 0.45 = \frac{91}{180}$ which is not steady.
I suspect there would be a solution if $\frac{1-0.6}{0.5}$ were equal to $\frac{1-0.55}{0.45}$ (or more sensibly, if the four numbers in the question were adjusted to make these equal), but they are not.