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I have a linearized time-delay system as follows:

$$frac{mathrm d X}{mathrm d t} = a[X(t)-X^*] + b [X(t-R) - X^*], $$
where $a$, $b$ are constants, $R$ is the constant delay, and $X^*$ is the equilibrium point.

How do I konw if, and under what condtion, this system is stable?

When $R=0$ or $b=0$ you can simply just look at whether $a+b<0$, but when $Rneq0$ and $bneq0$ the stability analysis requires some more effort.

You can formulate the stability analysis by assuming a solution of the following form

$$
X(t) = C,e^{lambda,t} + X^*.
$$

Plugging this into the left and right hand sides of the differential equations gives

begin{align}
frac{mathrm d X}{mathrm d t} &= C,lambda,e^{lambda,t}, \
a[X(t)-X^*] + b [X(t-R) - X^*] &= a,C,e^{lambda,t} + b,C,e^{lambda,(t - R)}.
end{align}

Equating these to each other and factoring out common terms gives

$$
left(lambda - a - b,e^{-lambda,R}right),C,e^{lambda,t} = 0.
$$

The trivial solution $C,e^{lambda,t} = 0$ of the above equation would imply $X(t) = X^*$, so would be stable. The other possible solutions require solving

$$
lambda = a + b,e^{-lambda,R},
$$

so the whole system would be stable if all of the solutions for $lambda$ have a negative real part. By using the substitution $lambda = tfrac{mu}{R} + a$ the above equation can also be written as

$$
mu,e^mu = b,R,e^{- a,R} = x,
$$

which has infinitely many solutions and can be found by using different branches of the Lambert W-function.

The stability condition that all of the solutions for $lambda$ must have a negative real part can also be expressed as that all solutions for $mu$ must have a real part larger than the real part of $-a,R$ when $R<0$ or smaller than the real part of $-a,R$ when $R>0$. If I remember correctly the principle branch of the Lambert W-function has the largest real part and there is no lower bound for the real part when considering all branches, but I couldn't find a source for this. So when $R>0$ the system is stable if the real part of the value of principal branch of the Lambert W-function evaluated at $x$ is smaller than the real part of $-a,R$ and when $R<0$ it will always be unstable because there will always be a branch which will violate the previously stated condition.

For example when assuming that $R>0$ and using a couple of values for $a,R$ and $b,R$ gives the following image, where blue means stable and yellow means unstable:

In the above image the real part of the value of principal branch of the Lambert W-function evaluated at $x$ plus $a,R$ is plotted. This value thus implies stability when it is negative.

The edge of stability can be found by setting the real part of $mu$ equal to $-a,R$, so it would be of the form $mu=-a,R+i,sigma$. Plugging this into the equation which was previously solved with the Lambert W-function gives

$$
(-a,R+i,sigma),e^{-a,R+i,sigma} = b,R,e^{- a,R}.
$$

Normally the real and imaginary part of the solution of the solution of the Lambert W-function is proportional to the branch number and the logarithm of the branch number respectively. So in order to give the solution of the principal branch $|sigma|$ should be small. When $sigma=0$ this gives $-a,R=b,R$, which matches with the edge in the image with $b,R>-1$. When $sigmaneq0$ it can be simplified to

$$
(-a,R+i,sigma),(cos(sigma)+isin(sigma)) = b,R.
$$

Solving this for $a,R$ and $b,R$ gives

begin{align}
a,R &= frac{sigma}{tan(sigma)}, \
b,R &= frac{-sigma}{sin(sigma)},
end{align}

which matches with the edge in the image with $b,R<-1$ when $sigmain[-pi,pi]$.

Another stability analysis tool that could be used is the Nyquist Criterion. Namely one could use the following transfer function as an open loop transfer function

$$
G(s) = frac{b,e^{-R,s}}{a - s}.
$$

The closed loop (and thus also your system) is stable if the Nyquist contour makes one counter clockwise encirclement around the minus one point when the real part of $a$ is positive, or when it makes no counter clockwise encirclements around the minus one point when the real part of $a$ is not positive.

For a given $a$ and $R$ it is then relatively easy to read for what values of $b$ the system will be stable. Namely $b$ is just a gain, so the gain margin could be used. The Nyquist contour can be obtained by evaluating $G(s)$ at $s=j,omega$ with $omegain[-infty,infty]$, which gives

$$
G(j,omega) = frac{b}{a^2 + omega^2} left(cos(R,omega) - j sin(R,omega)right) left(a + j,omegaright).
$$

The points of interest of the stability analysis are the places where the the Nyquist contour crosses the real axis, so when the imaginary part of $G(j,omega)$ is zero. At these frequencies of real axis crossings the stability edge cases will be at some of the cases where the real part equals minus one. Solving for this gives

begin{align}
a,R &= frac{R,omega}{tan(R,omega)}, \
b,R &= frac{-R,omega}{sin(R,omega)}.
end{align}

For $R,omegain[-pi,pi]$ this again gives the same stability edge with $b,R<-1$ in the image of the previous derivation. However, I did not find a way of using Nyquist to also show the same stability edge with $b,R>-1$.

After replacing $lambda$ with $lambda/R$, $a$ with $a/R$ and $b$ with $b/R$, it may be assumed that $R = 1$.

Set $u(t) = X(t) - X^ast$, write down the equation for $u$, and try $u(t) = e^{lambda t}$. This results in the characteristic equation
$$
lambda = a + be^{-lambda}
$$
The system is stable if this transcendental equation has only solutions with negative real parts. A sufficient condition is $a + |b| < 0$.

To find a relation between $a$ and $b$ that separates stable and unstable solutions, consider a purely imaginary $lambda = i omega$. The real and imaginary parts of the equation then become
$$
0 = a + b cos omega, quad omega = - b sin omega
$$

implying $tan omega = frac{omega}{a}$ and $ b = - frac{omega}{sin omega} = - a sec omega$.

You can now find marginally stable solutions for any $a$ by solving the first equation for $omega$ and then finding $b$ from the second equation.