$$\exists !xP(x) \equiv \exists x(P(x) \land \forall y(P(y) \rightarrow y=x))$$
By writing down what each of the side means in English sentence, I can show that both of them are are the same thing.
LHS: There exists unique value of x such that it satisfies P(x)
RHS: There exists some values x which satisfy P(x) and if any other value also satisfies P, then it is equal to x. This essentially means that all x are the same value. So it is unique.
But I'm unable to figure out a way of going to the RHS when only the LHS is given. How do I find the RHS when only the LHS is given?
Yes, you have done it.
You have taken the LHS. $~\exists! x~P(x)$
You have interpreted this to be equivalent to the statement, "There exists unique value of $x$ such that it satisfies $P(x)$".
You have established that, "There exists some values $x$ which satisfy $P(x)$ and/but if any value satisfies $P$, then it is equal to $x$," has an equivalent sematic meaning.
Then you translated that semantics into the syntax: $\exists x~(P(x)\wedge \forall y~(P(y)\to y=x))$
Then by the chain rule for equivalence, you are done. $$\exists!x~P(x)~\iff~\exists x~(P(x)\wedge \forall y~(P(y)\to y=x))$$
That is all that is required. It is very much the reason why the $\exists!$ quantifier is defined that way.